Trigonometry
Trigonometric equation using exact values
MJMT_Full_Test_05
Grade 12
Question:
The value of $x \in \left(0,\dfrac{\pi}{2}\right)$ satisfying $\dfrac{\sqrt{5}-1}{\sin x} + \dfrac{\sqrt{10+2\sqrt{5}}}{\cos x} = 8$ is
$\dfrac{\pi}{40}$
$\dfrac{3\pi}{10}$
$\dfrac{2\pi}{10}$
$\dfrac{4\pi}{10}$
Step-by-Step Solution
Key Concept: Recognise $\sqrt{5}-1=4\sin 18°$ and $\sqrt{10+2\sqrt{5}}=4\cos 18°$. The equation becomes $\frac{4\sin 18°}{\sin x}+\frac{4\cos 18°}{\cos x}=8$, i.e., $\sin(x+18°)=\sin x\cos x$.
$\sin(x+18°)=\sin 2x$: case $2x+x+18°=\pi$ gives $x=54°=3\pi/10$.
Correct Answer: 2