The number of solutions of the trigonometric | vedclass

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(C) Given equation: $1+\cos x \cdot \cos 5 x=\sin ^2 x$ Using the identity $\sin ^2 x = 1 - \cos ^2 x$,we get: $1+\cos x \cdot \cos 5 x = 1 - \cos ^2

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given equation: $1+\cos x \cdot \cos 5 x=\sin ^2 x$ Using the identity $\sin ^2 x = 1 - \cos ^2 x$,we get: $1+\cos x \cdot \cos 5 x = 1 - \cos ^2 x$ $\Rightarrow \cos ^2 x + \cos x \cdot \cos 5 x = 0$ $\Rightarrow \cos x(\cos x + \cos 5 x) = 0$ Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$: $\Rightarrow \cos x [2 \cos(3x) \cos(-2x)] = 0$ Since $\cos(-2x) = \cos(2x)$,we have: $2 \cos x \cos 3x \cos 2x = 0$ This implies $\cos x = 0$ or $\cos 3x = 0$ or $\cos 2x = 0$. For $x \in [0, 2 \pi]$: $1$. $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3 \pi}{2}$ ($2$ solutions) $2$. $\cos 3x = 0$ $\Rightarrow 3x = \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \frac{9 \pi}{2}, \frac{11 \pi}{2}$ $\Rightarrow x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}$ ($6$ solutions) $3$. $\cos 2x = 0$ $\Rightarrow 2x = \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}$ $\Rightarrow x = \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ ($4$ solutions) Combining these and removing duplicates $(\frac{\pi}{2}, \frac{3 \pi}{2})$,the distinct solutions are: $\{\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{5 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}, \frac{11 \pi}{6}\}$ Total number of solutions is $10$.

The number of solutions of the trigonometric equation $1+\cos x \cdot \cos 5 x=\sin ^2 x$ in $[0, 2 \pi]$ is

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