The solution set of the equation cos^2 2x + s | vedclass

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(C) Given equation: $\cos^2 2x + \sin^2 3x = 1$ $\Rightarrow \sin^2 3x = 1 - \cos^2 2x$ $\Rightarrow \sin^2 3x = \sin^2 2x$ $\Rightarrow \sin^2 3x - \

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$\left\{x \mid x = \frac{n\pi}{5}, n \in \mathbb{Z}\right\}$

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(C) Given equation: $\cos^2 2x + \sin^2 3x = 1$ $\Rightarrow \sin^2 3x = 1 - \cos^2 2x$ $\Rightarrow \sin^2 3x = \sin^2 2x$ $\Rightarrow \sin^2 3x - \sin^2 2x = 0$ Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$: $\Rightarrow \sin(3x + 2x) \sin(3x - 2x) = 0$ $\Rightarrow \sin 5x \sin x = 0$ Case $1$: $\sin 5x = 0$ $\Rightarrow 5x = n\pi$ $\Rightarrow x = \frac{n\pi}{5}, n \in \mathbb{Z}$ Case $2$: $\sin x = 0 \Rightarrow x = n\pi, n \in \mathbb{Z}$ Since $n\pi$ is a subset of $\frac{n\pi}{5}$ (by taking $n$ as a multiple of $5$),the general solution is $x = \frac{n\pi}{5}, n \in \mathbb{Z}$.

The solution set of the equation $\cos^2 2x + \sin^2 3x = 1$ is

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