Percubaan Matematik Tambahan SPM 2021 (Selangor), Kertas 1 (Soalan 14)

Soalan 14:
(a) Selesaikan persamaan 3 sin⁡ 2x = 4 kos⁡x untuk 0^∘ ≤ x ≤ 360^∘. [4 markah]
(b) Selesaikan persamaan serentak 2 kos⁡ (x – y) = √3 and 2 kos ⁡(x + y) = 1, dengan keadaan kedua-dua x dan y ialah sudut tirus. [4 markah]

Penyelesaian:
(a)

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


(b)

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