JEE Main 2021 — Trigonometric Equations Question with Solution

We know that $0\le {\sin }^{2}x\le 1$ and $0\le {\cos }^{2}x\le 1$

Also, we know for a number which lie between $0$ to $1,$ higher the power of the number the less is its value.  

$\Rightarrow {\sin }^{7}x\le {\sin }^{2}x\le 1 \dots \left(1\right)$

and ${\cos }^{7}x\le {\cos }^{2}x\le 1 \dots \left(2\right)$

Also, we know that ${\sin }^{2}x+{\cos }^{2}x=1$

This means the equality must hold for $\left(1\right)$ and $\left(2\right)$

$\Rightarrow {\sin }^{7}x={\sin }^{2}x$ and ${\cos }^{7}x={\cos }^{2}x$

$\Rightarrow {\sin }^{2}x\left({\sin }^{5}x-1\right)=0$ and ${\cos }^{2}x\left({\cos }^{5}x-1\right)=0$

$\Rightarrow \sin x=0$ or $\sin x=1$ and $\cos x=0$ or $\cos x=1$$\Rightarrow x=0, 2\pi , 4\pi , \frac{\pi }{2}, \frac{5\pi }{2}.$

Thus, there are total $5$ solutions.