The number of solutions to the equation cos^4 | vedclass

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(B) Given equation: $\cos^4 x + \frac{1}{\cos^2 x} = \sin^4 x + \frac{1}{\sin^2 x}$Rearranging the terms: $\cos^4 x - \sin^4 x = \frac{1}{\sin^2 x} -

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$4$

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(B) Given equation: $\cos^4 x + \frac{1}{\cos^2 x} = \sin^4 x + \frac{1}{\sin^2 x}$Rearranging the terms: $\cos^4 x - \sin^4 x = \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x}$Using the identity $a^2 - b^2 = (a-b)(a+b)$: $(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}$Since $\cos^2 x + \sin^2 x = 1$,we have: $\cos^2 x - \sin^2 x = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}$This implies $(\cos^2 x - \sin^2 x) \left(1 - \frac{1}{\sin^2 x \cos^2 x}\right) = 0$Using $\sin 2x = 2 \sin x \cos x$,we get $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$.So,$\cos 2x \left(1 - \frac{4}{\sin^2 2x}\right) = 0$This gives $\cos 2x = 0$ or $\sin^2 2x = 4$. Since $\sin^2 2x$ cannot be $4$,we must have $\cos 2x = 0$.Thus,$2x = (2n+1)\frac{\pi}{2}$,which means $x = (2n+1)\frac{\pi}{4}$.For $x \in [0, 2\pi]$,the solutions are $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.Therefore,the total number of solutions is $4$.

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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