Solve, for a given domain, trigonometric — Paper 2 · IGCSE Mathematics - Additional 0606

10 Solve, for 0° G x G 360° , the equation 3 (i) cot ( 2x - 10 °) = , [4] 4 (ii) sin 2 x - cos 2 x = cos x . [5]

9 marks

This question in 0606/22 Feb/March 2017

10 Solve the equation J r N r - = 3 for 0 G x G radians, [4] (i) 4 sin KK3x OO 4 2 L P (ii) 2 tan 2 y + sec 2 y = 14 sec y + 3 for 0° G y G 360° . [5]

9 marks

Mark scheme: 10(i) −1 3  M1 implied by 0.848[06…] sin   soi  4  0.848[06…] rot to 3 or more figs or M1 implied by a correct answer of acceptable 2.29[35…] rot to 3 or more figs accuracy 0.544 486... rot to 3 or more figs isw A1 1.03 or 1.02630... rot to 4 or more figs isw A1 Maximum 3 marks if extra angles in range; no penalty for extra values outside π range 0 ≤ x ≤ 2 10(ii) Correctly uses tan 2 y = sec 2 y − 1 and/or M1 for using correct relationship(s) to find an equation in terms of a single trigonometric sin y 2 2 and sin y = 1 − cos y ratio cos y 3sec 2 y − 14sec y − 5 = 0 DM1 for factorising or solving their 3-term quadratic dependent on the first M1 being ⇒ ( 3sec y + 1)( sec y − 5 ) awarded or 5cos 2 y + 14cos y − 3 = 0 ⇒ (5cos y − 1)(cos y + 3) 1 A1 [ cos y = −]3 cos y = 5 78.5 or 78.4630... rot to 2 or more decimal places A1 isw 281.5 or 281.536 … rot to 2 or more decimal places A1 Maximum 4 marks if extra angles in isw range; no penalty for extra values outside range0 ≤ x ≤ 360

This question in 0606/22 May/June 2017

10 Solve the equation (a) 2 sinx = 1 for - r G x G r radians, [3] (b) 3 tan ( 2y + 15c) = 1 for 0c G y G 180c, [4] (c) 3 cot 2 z = cosec 2 z - 7 cosec z + 1 for 0c G z G 360c. [5]

12 marks

Mark scheme: 10(a) sin x = 0.5 , sin x = − 0.5 M1 π π 5π 5π A2 A1 for any correct pair of angles , − , , − oe 6 6 6 6 if M0 then SC1 for a correct pair of angles 10(b) −1 1  M1 2 y + 15 = tan   soi  3  18.43(49...) and 198.43(49...) M1 1.7, 91.7 A2 A1 for each 10(c) Uses cot 2 z = cosec 2 z − 1 oe M1 for using correct identity or identities to obtain an equation in terms of a single trigonometric ratio 2cosec 2 z + 7cosec z − 4 = 0 ⇒ DM1 for dealing with quadratic ( 2cosec z − 1)( cosec z + 4 ) 1 M1 [sin z = 2 ] sin z = − 4 194.5, 345.5 A2 A1 for each

This question in 0606/23 May/June 2017

10 (i) Without using a calculator, solve the equation 6c 3 - 7c 2 + 1 = 0 . [5] It is given that y = tan x + 6 sin x . dy (ii) Find . [2] dx dy 3 2(iii) If = 7 show that 6 cos x - 7 cos x + 1 = 0 . [2] dx dy (iv) Hence solve the equation = 7 for 0 G x G r radians. [2] dx

11 marks

Mark scheme: 10(i) 3 2 B1 Or correct division. Finding or c = 1 → 6 (1) − 7 (1) + 1 = 0 → ( c − 1) is a using one correct factor. factor. Attempt to factorise or use long division to M1 obtain 6 c 2 …± 1 or 6c 2 ± c … respectively 6 c − c − 1 = 0 ( c − 1)( 2 A1 ) ( c − 1)( 2 c − 1)( 3c + 1) =0 A1 1 1 A1 FT c = 1, , − From three different linear factors 2 3 10(ii) dy 2 B2 B1 for each term = sec x + 6cosx dx 10(iii) 1 B1 B1dep + 6cosx = 7 cos 2 x 2 1 Replaces sec x by cos 2 x → 6cos 3 x − 7cos 2 x + 1 = 0 B1 B1dep Answer given so all steps must be correct. 10(iv) 1 1 A2 A1 for 2 values awrt cosx = 1, , − . A1 for third value and no others in 2 3 range. No credit for answers in  π  → x = 0, 1.05  or  , 1.91 degrees  3 

This question in 0606/22 Oct/Nov 2017

7 A particle moving in a straight line passes through a fixed point O. Its velocity, v ms -1 , t s after passing through O, is given by v = 3 cos 2t - 1 for t H 0 . (i) Find the value of t when the particle is first at rest. [2] r (ii) Find the displacement from O of the particle when t = . [3] 4 (iii) Find the acceleration of the particle when it is first at rest. [3]

8 marks

Mark scheme: 7(i) 1 M1 set v = 0 and solve for cos2t v = 0 → cos2t = 3 →=t 0.615 or 0.616 A1 7(ii) 3 M1A1 M1 for sin2t and ± t s = sin2t − t ( + c ) 2 π π A1 t = → s = 1.5 − ( = 0.715 ) 4 4 7(iii) a = −6sin2t M1A1 M1 for −sin2t t = 0.615 → a = –5.66 or –5.65 or −2 8 A1 condone substitution of degrees

This question in 0606/23 Oct/Nov 2017

10 (a) Show that + = 2 cosec x . [3] 1 + cos x sin x (b) Solve the following equations. (i) cot 2 y + cosec y - 5 = 0 for 0° G y G 360° [5] r 3(ii) cos 2z + =- for 0 G z G r radians [4] ` 4 j 2 Question 11 is printed on the next page.

12 marks

Mark scheme: 10(a) 2 2 B1 correct addition of fractions sin x + (1 + cos x ) LHS = sin x (1 + cos x ) 1 + 2cos x + 1 B1 expansion and use of identity = sin x (1 + cosx ) 2 (1 + cos x ) B1 factorisation and completion = = 2cosecx sinx (1 + cosx ) 10(b)(i) cosec 2 y −+1 cosecy − 5 = 0 M1 use of identity for cot2y to obtain quadratic in cosecy cosec2y + cosecy – 6 = 0 ( cosecy − 2 )( cosecy + 3 ) = 0 M1 solve 3 term quadratic for cosecy 1 1 M1 obtain values for siny sin y = , sin y = − 2 3 y = 30°, 150°, 199.5 °, 340.5 ° A2 A1 for 2 values 10(b)(ii) π 5π 7π M2 5π 2 z + = or (2.6…, 3.6…) M1 equate to 4 6 6 6 7π M1 equate to 6 7π 11π A2 A1 for 1 value z = or (0.916, 1.44) 24 24

This question in 0606/23 Oct/Nov 2017

11 (a) Solve 10 cos 2 x + 3 sin x = 9 for 0° 1 x 1 360° . [5] (b) Solve 3 tan 2y = 4 sin 2 y for 0 1 y 1 r radians. [5]

10 marks

Mark scheme: 11(a) 10 (1 − sin 2 x ) + 3sin x = 9 M1 Solves 10sin 2 x − 3sin x −=1 0 oe M1 dep on first M1 Solves their three term quadratic in sin ݔ 1 1 A1 sin x = , sin x = − 2 5 30º, 150º and 191.5º, 348.5º awrt A2 A1 for any two correct solutions 11(b) sin 2 y M1 3 = 4sin 2 y oe cos2 y Solves 3sin 2 y − 4sin 2 y cos2 y [ = 0] M1 dep on first M1 3 A1 sin2 y = 0 cos2 y = 4 Any two of A1 π, 0.72273…, 5.56045… nfww π A1 SC: cancels out sin2y after M1M0 , 0.361, 2.78 awrt nfww allow SC1 for 0.72273… and 5.56045... and 2 SC1for 0.361 and 2.78

This question in 0606/21 May/June 2018

11 (a) Solve 10 cos 2 x + 3 sin x = 9 for 0° 1 x 1 360° . [5] (b) Solve 3 tan 2y = 4 sin 2 y for 0 1 y 1 r radians. [5]

10 marks

Mark scheme: 11(a) 10 (1 − sin 2 x ) + 3sin x = 9 M1 Solves 10sin 2 x − 3sin x −=1 0 oe M1 dep on first M1 Solves their three term quadratic in sin ݔ 1 1 A1 sin x = , sin x = − 2 5 30º, 150º and 191.5º, 348.5º awrt A2 A1 for any two correct solutions 11(b) sin 2 y M1 3 = 4sin 2 y oe cos2 y Solves 3sin 2 y − 4sin 2 y cos2 y [ = 0] M1 dep on first M1 3 A1 sin2 y = 0 cos2 y = 4 Any two of A1 π, 0.72273…, 5.56045… nfww π A1 SC: cancels out sin2y after M1M0 , 0.361, 2.78 awrt nfww allow SC1 for 0.72273… and 5.56045... and 2 SC1for 0.361 and 2.78

This question in 0606/23 May/June 2018

7 (i) Show that - = 2 cosec x cot x . [4] 1 - cos x 1 + cos x 1 1 (ii) Hence solve the equation - = sec x for 0 G x G 2 r radians. [4] 1 - cos x 1 + cos x

8 marks

Mark scheme: 7(i) (1 + cos x ) − (1 − cos x ) M1 Taking common denominator (1 − cos x )(1 + cos x ) 2cosx A1 = 2 1 − cos x 2cosx M1 Using 1 − cos 2 x = sin 2 x = 2 sin x 2cosx 1 A1 Fully correct completion = × AG sinx sinx = 2cosecxcot x 7(ii) 2cosecxcotx = secx M1 2 1 A1 cot x = 2 0.955, 2.19, 4.10, 5.33 A2 A1 for 2 correct values A1 for further 2 correct values

This question in 0606/21 Oct/Nov 2018

8 (i) Show that - = 2 tan x sec x . [4] 1 - sin x 1 + sin x 1 1 (ii) Hence solve the equation - = cosec x for 0° G x G 360° . [4] 1 - sin x 1 + sin x

8 marks

Mark scheme: 8(i) (1 + sinx ) − (1 − sinx ) M1 (1 − sinx )(1 + sinx ) 2sin x A1 1 − sin 2 x 2sinx M1 cos 2 x 2sinx 1 A1 AG × = 2tan x secx cosx cosx 8(ii) M1 equate 2sec x tan x = cosecx 2 1 A1 tan x = 2 35.3°,144.7°, 215.3°, 324.7° 2 A1 two correct

This question in 0606/22 Oct/Nov 2018

9 (a) Solve 2 sin x + r = 3 for 0 1 x 1 r radians. [3] 4 (b) Solve 3 sec y = 4 cosec y for 0° 1 y 1 360° . [3]

12 marks

Mark scheme: 9(a) π π M1 x + = 4 3 π 5π A2 A1 for one correct and (0.262 and 1.31) 12 12 9(b) 1 1 M1 correctly use sec y = and cosec y = cos y sin y 4 A1 obtain expression for tany or y tan y = explicitly 3 53.1° and 233.1° A1 9(c) correctly rewrite equation in terms of sinz and cosz M1 use sin 2 z = 1 − cos 2 z M1 appropriate use of pythagorean identity for forming an equation in one trig ratio 8cos 2 z − 2cos z −=1 0 oe A1 ( 4cos z + 1)( 2cos z − 1) = 0 M1 solve 3 term quadratic in cosz 60° and 300° and 104.5° and 255.5° A2 A1 for any two correct

This question in 0606/23 Oct/Nov 2018

11 (a) (i) Show that i i = . [4] sin i 1 + cos i cosec i - cot i 5 (ii) Hence solve = for 180° 1 i 1 360 ° . [2] sin i 2 1 r(b) Solve tan 3z- 4 =- for 0 G z G radians. [3] ` j 2 2

9 marks

Mark scheme: 11(a)(i) 1  1 cosθ M2 M1 for either −   cosecθ− cotθ 1  cosθ sinθ  sinθ sinθ  = cosecθ−   sinθ sinθ sinθ  cosecθ− cotθ 1  1  or =  − cotθ sinθ sinθ sinθ  1 − cosθ M1 1 − cos 2 θ 1 − cosθ 1 A1 = (1 − cosθ)(1 + cosθ) 1 + cosθ 11(a)(ii) awrt 233.1 B2 with no extras in range 3 B1 for cosθ=− soi 5 11(b) −1 1  M1 3φ− 4 = tan  −  soi  2  awrt 0.132, 1.18 A2 with no extras in range A1 for one correct

This question in 0606/21 May/June 2019

2 (i) Show that - = cosecx . [3] 1 - cos x cosecx - cot x (ii) Hence solve = 2 for 0° 1 x 1 180° . [2] 1 - cos x

5 marks

Mark scheme: 2(i) sin 1 − cossin xx M1 express in terms of sinx and cosx 1 − cos x (1 − cos x ) A1 rewrite not as a fraction within a fraction sin (1 − cos x ) 1 A1 correct completion = cosec x answer given sin x 2(ii)  1  B1 sin x = x = 30°    2  x = 150º nfww B1 no extra answers

This question in 0606/23 Oct/Nov 2019

5 (a) Solve 3 cot 2 y - r = 1 for 0 1 y 1 r radians. [4] 4 (b) Solve 7 cot z + tan z = 7 cosec z for 0° G z G 360° . [6]

10 marks

Mark scheme: 5(a)  π  M1 ± 1.73… 3 tan  y −  = ( ± )  4  π π 2 π A1 1.04(7…) or 2.09(4…) y − = or 4 3 3 7 π A1 y = or 1.83 12 11π A1 y = or 2.88 12 5(b) correctly rewrite equation in terms of sinz M1 and cosz use sin2z = 1 – cos2z M1 appropriate use of Pythagorean identity for forming an equation in one trig ratio 6cos2z – 7cosz + 1 = 0 oe A1 (6cosz – 1) (cosz – 1) = 0 M1 solve three term quadratic in cosz 80.4º A1 279.6º A1

This question in 0606/23 Oct/Nov 2019

8 (a) Solve 3 cot 2 x - 14 cosec x - 2 = 0 for 0° 1 x 1 360°. [5] sin 4 y - cos 4 y (b) Show that = tan y - 2 cos y sin y. [4] cot y

9 marks

Mark scheme: 8(a) 3(cosec 2 x − 1) − 14cosec x − 2 [ = 0 ] M1 3cosec2 x − 14cosec x −=5 0 A1 (cosecx – 5)(3cosecx + 1) M1 1 A1 sinx = nfww 5 11.5 and 168.5 nfww A1 8(b) Correct use of sin 2 y + cos 2 y = 1 B1 Factorises using the difference of 2 B1 squares 1 cos y B1 Uses = tan y or cot y = cot y sin y correctly Full and correct completion to given B1 answer: tan y − 2cos y sin y

This question in 0606/22 May/June 2020

10 Solve the equation (a) 5 sec 2 A + 14 tan A - 8 = 0 for 0° G A G 180°, [4] (b) 5 sin 4B - r + 2 = 0 for - r G B G r radians. [4] b 8 l 4 4

8 marks

Mark scheme: 10(a) 5(1 + tan2A) + 14 tan A – 8 = 0 soi B1 Solves or factorises their 3-term quadratic M1 in tanA oe 11.3 and 108.4 A2 with no extras in range; or not from clearly wrong working but 11.30[99...] and 108.43[49...] rot to four or allow recovery from minor slips more decimal places or A1 for either, ignoring extras 10(b) π −1 2  B1 4 B − = sin  −  soi 8  5  −0.411[516...] rot to three or more figs M1 −0.00470[444...] A1 rot to three or more figs −0.584[344...] rot to three or more figs A1

This question in 0606/23 May/June 2020

4 It is given that y = ln ( sin x + 3 cos x) for 0 1 x 1 r. 2 dy (a) Find . [3] dx dy 1 (b) Find the value of x for which = - . [3] dx 2

6 marks

Mark scheme: 4(a) d y cos x − 3sin x 3 M1 for attempt at chain rule must have = function in numerator and denominator d x sin x + 3cos x A1 for denominator A1 for numerator (b) –2 cos x – 3 cos x = sin x – 6 sin x M1 Expand and collect terms in sin x and cos x 1 = tan x M1 sinx Use = tanx cos x π A1 Must be radians x = 4

This question in 0606/21 Oct/Nov 2020

(b) Solve the equation 5 tan x - 3 cot x = 2 sec x for 0° G x G 360 ° . [6]

10 marks

Mark scheme: 11(a) sinx M1 sinx sinx × Uses tanx = cosx cosx LHS = 1 − cosx 1 − cos 2 x M1 Dep Uses sin 2 x = 1 − cos 2 x to eliminate = cos x (1 − cosx ) sinx (1 − cos x )(1 + cosx ) 1 + cosx 2 M1Dep Factorise correctly and cancel = = secx + 1 correctly. cos x (1 − cosx ) cosx 1 A1 Uses = secx cosx 11(b) sinx cosx 2 B1 Change tan x, cot x and sec x into sin x and 5 − 3 = cos x correctly. cosx sinx cosx 2 2 M1 Multiply correctly by sin x cos x and use 5sin x − 3 1 − sin x = 2sin x ( ) 2 2 cos x + sin x = 1 8sin 2 x − 2sin x − 3 = 0 A1 Three term quadratic. ( 2sinx + 1)( 4sinx − 3 ) = 0 M1 Factorise or use formula on their quadratic 1 A1 sinx = − → x = 210° , 330° 2 3 A1 sinx = → x = 48.6°,131.4° 4

This question in 0606/22 Oct/Nov 2020

4 It is given that y = ln ( 1 + sin x) for 0 1 x 1 r. d y. [2] (a) Find d x d y r 1(b) Find the value of when x = , giving your answer in the form , where a is an integer. d x 6 a [2] d y (c) Find the values of x for which = tan x . [5] d x

9 marks

Mark scheme: 4(a) dy 1 M1 = dx 1 + sinx cosx A1 × cosx = 1 + sinx 4(b) π dy M1 insert into their 6 dx 1 A1 3 not 3 3 4(c) cos x sin x M1 sinx their = replace tan x with 1 + sin x cos x cos x use cos 2 x = 1 − sin 2 x M1 earned when equation reduced to a 2 quadratic in sinx 2sin x + sin x −=1 0 ( ) ( 2sin x − 1)( sin x + 1) = 0 M1 solve three term quadratic in sinx π A1 or 0.524 or better radians only x = 6 if M0 M0 M0 and (a) and (b) correct, allow SC2 for 1 π tanx = , x = 3 6 5 π A1 or 2.62 or better radians only x = A0 if extra solution(s) in range 6

This question in 0606/23 Oct/Nov 2020

5 It is given that y = 3 tan 2 x for 0° 1 x 1 360° . dy 2 (a) Show that = m tan x sec x where m is an integer to be found. [2] dx dy (b) Find all values of x such that = 3 sec x cosec x . [5] dx

7 marks

Mark scheme: 5(a) dy 2 B2 B1 for = 6tan x sec x dx d 2 1 2 (tan x ) = 2(tan x ) sec x dx 5(b) 6 tan x sec 2 x − 3sec x cosec x = 0 B1 NB division by secx is B0 3sec x (2 tan x sec x − cosec x ) = 0 oe 2 tan 2 x = 1 oe B1 1 M1 FT tan 2 x = k where k > 0 tan x = [ ± ] or [±] 0.707[1…] 2 35.3 or 35.2643… rot to 2 or more dp A2 no extras in range 215.3 or 215.2643… rot to 2 or more dp 144.7 or 144.7356… rot to 2 or more dp A1 for any two correct answers 324.7 or 324.7356… rot to 2 or more dp

This question in 0606/21 Oct/Nov 2021

3 (a) Show that + = 2 cot x cosec x . [4] sec x - 1 sec x + 1 1 1 (b) Hence solve the equation + = 3 sec x for 0° 1 x 1 360° . [4] sec x - 1 sec x + 1

8 marks

Mark scheme: 3(a) cos x cos x s ecx + 1 + sec x − 1 M1 + or 1 − cos x 1 + cos x sec 2 x − 1 cosx + cos 2 x + cos x − cos 2 x 2sec x A1 or 1 − cos 2 x tan 2 x 2cos x 2cos 2 x A1 or oe sin 2 x cos x sin 2 x Fully correct justification of given answer: 2cot xcosec x A1 3(b) 3tan 2 x = 2 oe or better, soi B1 or 5cos 2 x = 3 oe or better, soi or 5sin 2 x = 2 oe or better, soi 2 M1 FT an equation of the form tanx = [ ± ] oe or [±] 0.816[4…] 2 a tan x = b , a > 0, b > 0 3 or p sin 2 x = q or p cos 2 x = q 3 or cosx = [ ± ] oe or [±] 0.774[5…] where p > 0, q > 0 and p > q 5 2 or sinx = [ ± ] oe or [±] 0.632[4…] 5 39.2° or 39.2315… rot to 2 or more dp A2 no extras in range 140.8° or 140.7684… rot to 2 or more dp 219.2° or 219.2315… rot to 2 or more dp 320.8° or 320.7684… rot to 2 or more dp A1 for any two correct answers

This question in 0606/22 Oct/Nov 2021

5 (a) Show that + = 2 tan x sec x . [4] cosec x - 1 cosec x + 1 1 1 (b) Hence solve the equation + = 5 cosec x for 0° 1 x 1 360° . [4] cosec x - 1 cosec x + 1

8 marks

Mark scheme: 5(a) sin x sin x cosec x + 1 + cosec x − 1 M1 + or oe 1 − sin x 1 + sin x cosec 2 x − 1 sin x + sin 2 x + sin x − sin 2 x 2cosec x A1 or oe 1 − sin 2 x cot 2 x 2sin x 2sin 2 x A1 or oe cos 2 x sin x cos 2 x Fully correct justification of given answer: A1 2sin x 1 × = 2tan x sec x cos x cos x 1 or 2tan x × = 2tan x sec x cos x 2sin x or × sec x = 2tan x sec x cos x or equivalent 5(b) 2tan 2 x = 5 or better, soi B1 or 7cos 2 x = 2 or better, soi or 7sin 2 x = 5 or better, soi 5 M1 FT an equation of the form tan x = [ ± ] oe or [±] 1.58[1…] 2 a tan x = b a > 0, b > 0 2 or p sin 2 x = q or p cos 2 x = q 2 or cos x = [ ± ] oe or [±] 0.534[5…] where p > 0, q > 0 and p > q 7 5 or sin x = [ ± ] oe or [±] 0.845[1…] 7 57.7 or 57.6884… rot to 2 or more dp A2 no extras in range 237.7 or 237.6884… rot to 2 or more dp A1 for any two correct answers 122.3 or 122.3115… rot to 2 or more dp 302.3 or 302.3115… rot to 2 or more dp

This question in 0606/23 Oct/Nov 2021

7 In this question, all angles are in radians. (a) Solve the equation sec 2i = tan i + 3 for - r 1 i 1 r . [5] r tan z (b) Show that, for 0 1 z 1 , = sec z . [3] 2 1 - cos 2z 17 3 r (c) Given that cosecx =- and that 1 x 1 2 r , find the exact value of cotx. [2] 8 2

10 marks

Mark scheme: 7(a) Uses a valid Pythagorean identity to write B1 in terms of a single trig ratio e.g. 1 + tan 2 θ = tanθ + 3 Rearranges and factorises/solves M1 e.g. tan 2 θ − tan θ − 2 = 0 (tanθ − 2)(tanθ + 1) = 0 tanθ = 2 tanθ = −1 soi A1 π 3π A2 A1 for any 3 correct, ignoring extras 1.11, −2.03, − , and no extras in 4 4 range cao 7(b) sin φ M1 Use of tan φ = in a correct cosφ expression or correct working Use of 1 − cos 2 φ = sin 2 φ in a correct M1 expression or correct working 1 A1 nfww Completion with = secφ cosφ 7(c) 2 M1 cot x = [ − ] cosec x − 1 soi 8 8 or sin x = − and tan x = − soi 17 15 8 15 or sin x = − and cos x = soi 17 17 1 or  −1  8   tan  sin  −     17   15 A1 − or −1.875 cao, isw 8

This question in 0606/22 Feb/March 2022

5 You are given that y = . cos 2x d y k sin 2x (a) Show that = 2 where k is a constant to be found. [2] d x cos 2x d y 5 r (b) Find the values of x such that = for 0 1 x 1 . [4] d x sin 2x 2

6 marks

Mark scheme: 5(a) dy −2 2sin2 x B2 B1 for − (cos2 x ) −2  m sin 2 x = −(cos2 x ) −2sin2 x = dx cos 2 2 x 0[cos2 x ] − ( m sin2 x ) or where m = 2 dy 0[cos2 x ] −−( 2sin2 x ) 2sin2 x 2 or = = cos 2 x dx cos 2 2 x cos 2 2 x or m < 0 5(b) 2tan22x = 5 M1 FT their k or 7cos22x = 2 or 7sin22x = 5 5 A1 tan2 x =  2 2 or cos2 x =  7 5 or sin2 x =  7 0.503 or 0.5034[26…] rot to 4 or more sf A2 A1 for either, ignoring extras 1.07 or 1.067[36…] rot to 4 or more sf and no extras in range

This question in 0606/21 Oct/Nov 2022

7 (a) Show that + = 2 cosec x . [4] 1 - cos x sin x sin x 1 - cos x (b) Hence solve the equation + = 3 sin x - 1 for 0° 1 x 1 360° . [4] 1 - cos x sin x

8 marks

Mark scheme: 7(a) sin 2 x + (1 − cos x ) 2 M1 (1 − cos x )sin x sin 2 x (1 − cos x) 2 or + (1 − cos x )sin x (1 − cos x)sin x sin 2 x + 1 − 2cos x + cos 2 x A1 1 − cos 2 x + (1 − cos x ) 2 OR (1 − cos x )sin x (1 − cos x )sin x 1 + 1 − 2cos x A1 (1 − cos x )(1 + cos x ) + (1 − cos x) 2 OR (1 − cos x )sin x (1 − cos x )sin x 1 − cos 2 x + 1 − 2cos x + cos 2 x or (1 − cos x )sin x Fully correct justification of given A1 All steps correct and final step justified answer: 2(1 − cos x ) 1 + cos x + 1 − cos x = 2cosec x OR = 2cosec x (1 − cos x )sin x sin x 2 − 2cos x 2 or = = 2cosec x (1 − cos x )sin x sin x or equivalent Alternative sin x (1 + cos x ) (1 − cos x )sin x (M1) + (1 − cos x )(1 + cos x ) sin x sin x sin x (1 + cos x ) (1 − cos x )sin x or + 1 − cos 2 x sin 2 x sin x + sin x cos x sin x − cos x sin x (A1) + sin 2 x sin 2 x 2sin x (A1) sin 2 x Fully correct justification of given (A1) All steps correct and final step justified 2 answer: = 2cosec x sin x 7(b) 3sin 2 x − sin x − 2  = 0 soi B1 (3sinx + 2)(sinx – 1) [= 0] oe M1 2 A1 sin x = − , sin x = 1 3 90 A1 and no extras in range 221.8 or 221.81[03…] rot to 2 or more If B1 M1 A0 A0 allow SC1 for dp 221.8 or 221.81[03…] rot to 2 or 318.2 or 318.18[96…] rot to 2 or more more dp dp and 318.2 or 318.18[96…] rot to 2 or more dp and no extras in range

This question in 0606/22 Oct/Nov 2022

5 (a) Show that + = 2 sec x . [4] 1 - sin x cos x i i cos 1 - sin 2 2 2 i (b) Hence solve the equation + = 8 cos for - 360° 1 i 1 360° . [4] i i 2 1 - sin cos 2 2

8 marks

Mark scheme: 5(a) cos 2 x + (1 − sin x ) 2 M1 Correctly takes common denominator (1 − sin x ) cos x cos 2 x (1 − sin x ) 2 or + (1 − sin x ) cos x (1 − sin x ) cos x cos 2 x + 1 − 2sin x + sin 2 x A1 1 − sin 2 x + (1 − sin x ) 2 OR (1 − sin x ) cos x (1 − sin x ) cos x 1 + 1 − 2sin x A1 (1 − sin x )(1 + sin x ) + (1 − sin x ) 2 (1 − sin x ) cos x OR (1 − sin x ) cos x 1 − sin 2 x + 1 − 2sin x + sin 2 x or (1 − sin x ) cos x 2(1 − sin x ) A1 All steps correct and final step justified = 2sec x (1 − sin x ) cos x 2 − 2sin x 2 or = = 2sec x 1 + sin x + 1 − sin x (1 − sin x ) cos x cosx OR = 2sec x cos x or equivalent Alternative Must work with LHS only (cos x )(1 + sin x ) (1 − sin x )cos x (M1) Forms fractions with common + (1 − sin x )(1 + sin x ) ( cos x ) cos x denominator in different form (cos x )(1 + sin x ) (1 − sin x )cos x (A1) Uses difference of two squares and + 2 cos x cos 2 x sin 2 x + cos 2 x = 1 to write fractions with a common denominator in the same form 2cos x (A1) Combine as a single fraction and 2 collects terms cos x 2 (A1) All steps correct and final step justified = 2sec x cos x 5(b) 3 1 B1 cos = 2 4 1 M1  2  cos = 3 their soi dep on starting with 2sec = 8cos 2 4 2 2 101.9 awrt A2 and no extras in range A1 for either, ignoring extras in range If A0 then SC1 for 102 with no extras in range

This question in 0606/23 Oct/Nov 2022

(b) Solve the equation 3 tan 2 ( y + r ) = 1 for - 2r 1 y 1 0 . [4] 4

9 marks

Mark scheme: 11(a) 1 1 M1 + 1 1 1 1 − + cos x sin x cos x sin x Simplifies denominator A1 1 1 + sin x − cos x sin x + cos x sin x cos x sin x cos x Writes as two simple algebraic fractions: A1 OR writes as a single simple algebraic fraction: sin x cos x sin x cos x sin x cos x (sin x + cos x ) + sin x cos x (sin x − cos x ) + sin x − cos x sin x + cos x (sin x − cos x )(sin x + cos x ) Combines and simplifies: A1 2sin 2 x cos x sin 2 x − cos 2 x Correct simplification to given answer e.g. A1 All steps correct nd fully justified 2sin 2 x cos x 2cos x = 2 cos 2 x 1 − cot 2 x sin x − sin 2 x sin 2 x (2cos x )  2cos x  or = 2 2  2  sin x (1 − cot x )  1 − cot x  11(a) Alternative method Common denominator: (M1) sec x + cosec x + sec x − cosec x (sec x − cosec x )(sec x + cosec x ) 2sec x (A1) Simplifies: sec 2 x − cosec 2 x Rewrites in terms of sinx and cosx: (A1) OR multiplies numerator and denominator by 2 cos2x: cos x 2sec x cos 2 x  1 1 2 2 2 − sec x − cosec x cos x cos 2 x sin 2 x 2 (A1) cos x cos 2 x  1 1 cos 2 x − cos 2 x sin 2 x Correct simplification to given answer e.g. (A1) 2cos 2 x cos x 2cos x = cos 2 x cos 2 x 1 − cot 2 x − cos 2 x sin 2 x 11(b)    B1 tan  y +  = 1  4  3  π     7  M1 y + = , or − , or − , or − oe    4  6 6 6 6  5 13 17  A2 No extras within range  y =  − , − , − , − oe 12 12 12 12 A1 for two correct, ignoring extras

This question in 0606/21 Oct/Nov 2023

10 (a) By writing cotx and tanx in terms of cosx and sinx, show that sin x cos x + = sin x + cos x. [5] 1 - cot x 1 - tan x (b) Solve the equation 9 cot x + 3 cosec x = tan x , for 0° 1 x 1 360° . [5]

10 marks

Mark scheme: 10(a) Writes cotx and tanx in terms of sinx and M1 OR cosx:  sin x   cos x  sin x  1 −  + cos x  1 −  sin x cos x  cos x   sin x  + cos x sin x  cos x  sin x  1 − 1 −  1 −  1 −  sin x cos x  sin x  cos x  Simplifies denominator: A1 OR sin x cos x  cos x − sin x   sin x − cos x  + sin x   + cos x   sin x − cos x cos x − sin x  cos x   sin x  sin x cos x  sin x − cos x  cos x − sin x      sin x  cos x  Writes as two simple algebraic fractions: A1 OR writes as a single simple algebraic sin 2 x cos 2 x fraction: + 2 2 sin x (cos x − sin x ) + cos x (sin x − cos x ) sin x − cos x cos x − sin x (sin x − cos x )(cos x − sin x ) Writes as a difference with a common A1 sin 2 x (cos x − sin x ) − cos 2 x (cos x − sin x ) OR denominator: (sin x − cos x )(cos x − sin x ) sin 2 x cos 2 x − sin x − cos x sin x − cos x Correct simplification to given answer, e.g., A1 All steps correct and final step fully justified (sin x − cos x )(sin x + cos x ) by factorising = sin x + cos x (sin x − cos x ) or (sin x − cos x ) (sin x + cos x )  = sin x + cos x  (sin x − cos x ) 10(b) 10cos 2 x + 3cos x − 1[ = 0] B2 9cos x 3 sin x B1 for + = or better or sec 2 x − 3sec x − 10[ = 0] sin x sin x cos x 3tan x 2 or 9 + = tan x or better sin x OR M1 for one sign error in 10cos 2 x + 3cos x − 1[ = 0] or sec 2 x − 3sec x − 10[ = 0] ( 5cos x − 1)( 2cos x + 1) = 0  M1 FT their 3-term quadratic in cosx or secx or ( sec x − 5 )( sec x + 2 ) = 0  1 1 A2 A1 for any two correct angles [cosx = and cosx = − 5 2 1 1 [found using cosx = and cosx = − 5 2 OR OR secx = 5 and secx = –2 leading to] secx = 5 and secx = –2]; 78.5 or 78.46[30…] rot to 2 or more dp ignore extras 281.5 or 281.53[69…] rot to 2 or more dp 120 240 and no extras in range 0  x  360

This question in 0606/22 Oct/Nov 2023

sin x cos x(b) Hence solve the equation - = 1 for 0° 1 x 1 360° . [5] tan x - 1 tan x + 1

10 marks

Mark scheme: 8(a) sin x cos x M1 sin x (tan x + 1) − cos x (tan x − 1) − OR sin x sin x (tan x − 1)(tan x + 1) − 1 + 1 cos x cos x sin x cos x A1 OR − sin x − cos x sin x + cos x sin x tan x + sin x − cos x tan x + cos x cos x cos x (tan x − 1)(tan x + 1)  sin x   sin x  sin x  + 1  − cos x  − 1   cos x   cos x   sin x   sin x  OR sin x  + 1  − cos x  − 1  2 sin x  cos x   cos x  − 1 OR cos 2 x  sin x  sin x   − 1  + 1   cos x  cos x  sin x cos x cos 2 x A1 sin 2 x − + sin x − sin x + cos x sin x − cos x sin x + cos x OR cos x  sin x + cos x   sin x − cos x  sin 2 x sin x   − cos x   2 − 1  cos x   cos x  cos x OR sin 2 x − cos 2 x cos 2 x sin 2 x cos x + sin x cos 2 x − cos 2 x sin x + cos 3 x A1 sin 2 x + cos 2 x sin 2 x + cos 2 x sin 2 x + cos xsinx − cosxsinx − cos 2 x cos x cos x OR or OR sin 2 x − cos 2 x 1 − 2cos 2 x sin 2 x + sin x cos x − cos x sin x + cos 2 x cos 2 x cos 2 x cos 2 x  cos x sin 2 x − cos 2 x Fully correct justification of given answer e.g. A1 All steps correct and final step justified 1 cos 2 x cos x cos x sin 2 x + cos 2 x ( ) cos x  = = oe 2 2 2 2 2 2 2 2 cos x sin x − cos x sin x − cos x sin x − cos x sin x − cos x oe 2 x − cos 2 x or8(b) 2cos 2 x + cos x − 1  = 0 B2 B1 for cos x = 1 − cos better ( 2cos x − 1)( cos x + 1) = 0  M1 FT their 3-term quadratic in cosx [ x =] 60,300,180 A2 A1 for any two correct , ignoring and no extras in range extras

This question in 0606/23 Oct/Nov 2023

10 (a) Show that ( tan x + sec x) 2 can be written as . [4] 1- sinx (b) Hence solve the equation ( tan 3 i + sec 3i ) 2 = 6 for 0° G i G 180 ° . [4]

8 marks

Mark scheme: 10(a) tan 2 x  2tan x sec x  sec 2 x M1  sin x 1  2 or     cos x cos x  sin 2 x sin x 1 1 A1  2    cos 2 x cos x cos x cos 2 x 1 2 or factorises sin x  1 oe 2  cos x (1  sin x ) 2 A1 1  sin 2 x (1  sin x )(1  sin x ) 1  sin x A1 must be fully justified  (1  sin x )(1  sin x ) 1  sin x or (1  sin x ) 2 1  sin x  (1  sin x )(1  sin x ) 1  sin x 10(b) 7sin3 5 B1 One correct value for 3 soi e.g. M1 45.58… 134.4… 405.5… 494.4… 15.2 or 15.19 to 15.195 A2 with no extras in range 44.8 or 44.80 to 44.81 135.2 or 135.19 to 135.195 A1 for any 2 correct, ignoring extras 164.8 or 164.80 to 164.81

This question in 0606/21 May/June 2024

- 1 x 1 .5 (a) The function f is defined by f ( x) = 2 for r r cos x 2 2 (i) Show that f ( x) can be written as a tan 2 x + b , where a and b are integers. [2] (ii) Hence solve the equation f ( x) = 4 . [3] (iii) Hence also find the gradient of the curve y = f ( x) at each of the points where y = 4 . [4] (b) Solve the equation 50 cos 2i = 5 sin i + 47 for 0° G i G 360 ° . [5]

14 marks

Mark scheme: 5(a)(i) sec 2 x  2tan 2 x oe and 2 M1 for use of a relationship to form a 2 convincing correct statement from correct completion to 3tan x  1 nfww which the answer can be easily 2 2sin 2 x determined e.g. sec x  or cos 2 x 1 2  2tan x cos 2 x 5(a)(ii) tan x 1 soi M1 4  their1 FT tan x  providing their 3 4  their1 > 0 their 3 π A2 A1 for each, ignoring extra solutions  x  π ,  or 0.785[39…] nfww 4 4 and no other solutions 5(a)(iii) f ( x )  6tan x sec 2 x oe M2 FT 2(their 3)tan x sec 2 x M1 for f ( x )  k tan x sec 2 x where k ≠ 2their 3   π     π   A2 A1 for each nfww  f      12;  f      12 nfww   4     4   5(b) Correct use of sin 2  cos 2  1 to form M1 Condone one sign or arithmetic error in rearrangement a 3-term quadratic in sinin solvable form 50sin 2  5sin 3   0  oe Solves or factorises their 3-term quadratic M1 FT their 3-term quadratic in sin in sine.g. (10sin+ 3)(5sin– 1) [= 0] sin= 0.3 sin= 0.2 soi A1 11.5 or 11.53[69...] A2 with no extras in range 168.5 or 168.46[30...] 197.5 or 197.45[76...] A1 for any two correct angles, ignoring 342.5 or 342.54[23...] extras

This question in 0606/23 May/June 2024