CAIE P1 2002 November — Question 5 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2002
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeShow then solve: tan/sin/cos identity manipulation
DifficultyModerate -0.3 This is a standard P1 trigonometric equation requiring routine manipulation (converting tan to sin/cos, using sin²θ + cos²θ = 1) followed by solving a quadratic equation and finding angles in a given range. The conversion is guided in part (i), making this slightly easier than average but still requiring multiple standard techniques.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

5
  1. Show that the equation \(3 \tan \theta = 2 \cos \theta\) can be expressed as $$2 \sin ^ { 2 } \theta + 3 \sin \theta - 2 = 0$$
  2. Hence solve the equation \(3 \tan \theta = 2 \cos \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

AnswerMarks Guidance
(i) \(3\tan^2\theta = 2\cos\theta\). \(3\sin^2\theta + \cos\theta - 2\cos\theta = 2(1 - \sin^2\theta)\). \(3s = 2(1 - s^2)\)M1, M1, A1 Use of \(t = s - c\); Use of \(s^2 + c^2 = 1\); Everything correct – answer given
(ii) Soln of \(2s^2 + 3s - 2 = 0\). \(s = 0.5\) or \(-2\). \(\theta = 30°\) or \(150°\)M1, A1 A1∨ Correct method of solution; Correct only, then \(\sqrt{\phantom{x}}\) for 180 – first answer or consistent with his cosine-loses \(\sqrt{\phantom{x}}\) mark if extra solutions
**(i)** $3\tan^2\theta = 2\cos\theta$. $3\sin^2\theta + \cos\theta - 2\cos\theta = 2(1 - \sin^2\theta)$. $3s = 2(1 - s^2)$ | M1, M1, A1 | Use of $t = s - c$; Use of $s^2 + c^2 = 1$; Everything correct – answer given

**(ii)** Soln of $2s^2 + 3s - 2 = 0$. $s = 0.5$ or $-2$. $\theta = 30°$ or $150°$ | M1, A1 A1∨ | Correct method of solution; Correct only, then $\sqrt{\phantom{x}}$ for 180 – first answer or consistent with his cosine-loses $\sqrt{\phantom{x}}$ mark if extra solutions
5 (i) Show that the equation $3 \tan \theta = 2 \cos \theta$ can be expressed as

$$2 \sin ^ { 2 } \theta + 3 \sin \theta - 2 = 0$$

(ii) Hence solve the equation $3 \tan \theta = 2 \cos \theta$, for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P1 2002 Q5 [6]}}