The number of solutions of the equation 1 + s | vedclass

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(C) Given the equation $1 + \sin^4 x = \cos^2 3x$.Since $\sin^4 x \ge 0$,the left side $1 + \sin^4 x \ge 1$.Since $\cos^2 3x \le 1$,the right side $\c

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the equation $1 + \sin^4 x = \cos^2 3x$.Since $\sin^4 x \ge 0$,the left side $1 + \sin^4 x \ge 1$.Since $\cos^2 3x \le 1$,the right side $\cos^2 3x \le 1$.For the equality to hold,we must have $1 + \sin^4 x = 1$ and $\cos^2 3x = 1$.$1 + \sin^4 x = 1 \implies \sin^4 x = 0 \implies \sin x = 0 \implies x = n\pi$ for $n \in \mathbb{Z}$.Now check $\cos^2 3x = 1$ for $x = n\pi$:$\cos^2(3n\pi) = (\pm 1)^2 = 1$. This is always true for any integer $n$.We need $x \in [-\frac{5\pi}{2}, \frac{5\pi}{2}]$,which is $[-2.5\pi, 2.5\pi]$.The possible values for $x = n\pi$ in this interval are:$n = -2 \implies x = -2\pi$$n = -1 \implies x = -\pi$$n = 0 \implies x = 0$$n = 1 \implies x = \pi$$n = 2 \implies x = 2\pi$The solutions are $\{-2\pi, -\pi, 0, \pi, 2\pi\}$.There are $5$ solutions.

The number of solutions of the equation $1 + \sin^4 x = \cos^2 3x$ for $x \in [-\frac{5\pi}{2}, \frac{5\pi}{2}]$ is

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