The number of real solutions x of the equatio | vedclass

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(B) Given the equation: $\cos^2(x \sin(2x)) + \frac{1}{1+x^2} = \cos^2 x + \sec^2 x$.Consider the Left Hand Side $(LHS)$: $f(x) = \cos^2(x \sin(2x)) +

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$1$

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(B) Given the equation: $\cos^2(x \sin(2x)) + \frac{1}{1+x^2} = \cos^2 x + \sec^2 x$.Consider the Left Hand Side $(LHS)$: $f(x) = \cos^2(x \sin(2x)) + \frac{1}{1+x^2}$.Since $\cos^2(	heta) \leq 1$ and $\frac{1}{1+x^2} \leq 1$ for all real $x$,we have $f(x) \leq 1 + 1 = 2$.Consider the Right Hand Side $(RHS)$: $g(x) = \cos^2 x + \sec^2 x$.By the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,$\cos^2 x + \sec^2 x \geq 2 \sqrt{\cos^2 x \cdot \sec^2 x} = 2 \sqrt{1} = 2$.For the equation to hold,we must have $f(x) = 2$ and $g(x) = 2$.$g(x) = 2$ occurs when $\cos^2 x = \sec^2 x$,which implies $\cos^4 x = 1$,so $\cos x = \pm 1$,meaning $x = n\pi$ for $n \in \mathbb{Z}$.If $x = 0$,$f(0) = \cos^2(0) + \frac{1}{1+0} = 1 + 1 = 2$ and $g(0) = \cos^2(0) + \sec^2(0) = 1 + 1 = 2$.If $x = n\pi$ where $n 
eq 0$,then $\frac{1}{1+x^2} Thus,the only real solution is $x = 0$.

The number of real solutions $x$ of the equation $\cos^2(x \sin(2x)) + \frac{1}{1+x^2} = \cos^2 x + \sec^2 x$ is

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