SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 14)

Question 14:
(a) Solve the equation 3 sin⁡ 2x = 4 cos⁡x for 0^∘ ≤ x ≤ 360^∘. [4 marks]
(b) Solve the simultaneous equations 2 cos⁡ (x – y) = √3 and 2 cos ⁡(x + y) = 1, where x and y are both acute angles. [4 marks]

Solution:
(a)

$\begin{array}{l}3\sin 2x=4\cos x\\ 3(2\sin x\cos x)=4\cos x\\ 6\sin x\cos x=4\cos x\\ 6\sin x\cos x-4\cos x=0\\ 2\cos x\left(3\sin x-2\right)=0\\ 2\cos x=0\\ \cos x=0\\ x={90}^o,{270}^o\\ \\ 3\sin x-2=0\\ 3\sin x=2\\ \sin x=\frac{2}{3}\\ x={\sin }^{-1}\frac{2}{3}\\ x={41.81}^o,\text{ (18}{0}^o-{41.81}^o)\\ x={41.81}^o,\text{ 138}{\text{.19}}^o\\ \\ \therefore x={41.81}^o,{90}^o,\text{ 138}{\text{.19}}^o,{270}^o\end{array}$


(b)

$\begin{array}{l}2\cos \left(x-y\right)=\sqrt{3}\\ \cos \left(x-y\right)=\frac{\sqrt{3}}{2}\\ \left(x-y\right)={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)\\ \left(x-y\right)=30\mathrm{\dots ..}\left(1\right)\\ \\ 2\cos \left(x+y\right)=1\\ \cos \left(x+y\right)=\frac{1}{2}\\ \left(x+y\right)={\cos }^{-1}\left(\frac{1}{2}\right)\\ \left(x+y\right)=60\mathrm{\dots ..}\left(2\right)\\ \\ x-y=30\mathrm{\dots ..}\left(1\right)\\ x+y=60\mathrm{\dots ..}\left(2\right)\\ \\ \left(1\right)+\left(2\right),\\ 2x=90\\ x=45\\ \\ \text{From }\left(1\right),\\ 45-y=30\\ y=15\end{array}$