The number of solutions to the equation cos ^ | vedclass

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Question

(b)

We have,

$\cos ^4 x+\frac{1}{\cos ^2 x}=\sin ^4 x+\frac{1}{\sin ^2 x}$

$\Rightarrow \quad \cos ^4 x-\sin ^4 x=\frac{1}{\sin ^2 x}-\frac{1

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(b)

We have,

$\cos ^4 x+\frac{1}{\cos ^2 x}=\sin ^4 x+\frac{1}{\sin ^2 x}$

$\Rightarrow \quad \cos ^4 x-\sin ^4 x=\frac{1}{\sin ^2 x}-\frac{1}{\cos ^2 x}$

$\Rightarrow \quad\left(\cos ^2 x-\sin ^2 xight)\left(\cos ^2 x+\sin ^2 xight)$

$=\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cos ^2 x}$

$\Rightarrow \quad \cos ^2 x-\sin ^2 x=\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cos ^2 x}$

$\Rightarrow \quad \cos 2 x\left(1-\frac{4}{\sin ^2 2 x}ight)=0$

$\Rightarrow \quad \cos 2 x=0$ or $\sin ^2 2 x=4$

$	herefore \quad \cos 2 x=0$ or $\sin ^2 2 x 
eq 4$

$\Rightarrow \quad 2 x=(2 n+1) \frac{\pi}{2}$

$\Rightarrow \quad x=(2 n+1) \frac{\pi}{4}$

$\ln x \in(0,2 \pi)$

$x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$

$	herefore$ Total number of solution $=4$

The number of solutions to the equation $\cos ^4 x+\frac{1}{\cos ^2 x}=\sin ^4 x+\frac{1}{\sin ^2 x}$ in the interval $[0,2 \pi]$ is

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