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Q.
If $\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$, then the two curves $y=\cos x$ and $y=\sin 3 x$ intersect at
Trigonometric Functions
Solution:
At the intersection point of $y=\cos x$ and $y=\sin 3 x$,
we have $\cos x=\sin 3 x$
$\Rightarrow \cos x=\cos \left(\frac{\pi}{2}-3 x\right)$
$\Rightarrow x=2 n \pi \pm\left(\frac{\pi}{2}-3 x\right)$
$\Rightarrow x=\frac{\pi}{4}, \frac{\pi}{8} [\because-\pi / 2 \leq x \leq \pi / 2]$
So, $y=\cos \frac{\pi}{4} \text { at } x=\frac{\pi}{4}$
and $y=\cos \frac{\pi}{8}, \text { at } x=\frac{\pi}{8}$
Thus, the points are $\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{\pi}{8}, \cos \frac{\pi}{8}\right)$