If the general solution of sin x + 3 sin 3x + | vedclass

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(D) Given equation: $\sin x + 3 \sin 3x + \sin 5x = 0$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:

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$\{-1, -\frac{1}{2}, \frac{1}{2}, 1\}$

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(D) Given equation: $\sin x + 3 \sin 3x + \sin 5x = 0$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$: $(\sin 5x + \sin x) + 3 \sin 3x = 0$ $2 \sin 3x \cos 2x + 3 \sin 3x = 0$ $\sin 3x (2 \cos 2x + 3) = 0$ Case $1$: $\sin 3x = 0$ $\Rightarrow 3x = n\pi$ $\Rightarrow x = \frac{n\pi}{3}$ Case $2$: $2 \cos 2x + 3 = 0 \Rightarrow \cos 2x = -\frac{3}{2}$ Since the range of $\cos 2x$ is $[-1, 1]$,$\cos 2x = -\frac{3}{2}$ has no real solution. Thus,$x = \frac{n\pi}{3}$. For $x = \frac{n\pi}{3}$,the possible values of $\cos x$ are $\cos(0) = 1$,$\cos(\frac{\pi}{3}) = \frac{1}{2}$,$\cos(\frac{2\pi}{3}) = -\frac{1}{2}$,$\cos(\pi) = -1$,$\cos(\frac{4\pi}{3}) = -\frac{1}{2}$,$\cos(\frac{5\pi}{3}) = \frac{1}{2}$. Therefore,the set of all values of $\cos y$ is $\{-1, -\frac{1}{2}, \frac{1}{2}, 1\}$.

If the general solution of $\sin x + 3 \sin 3x + \sin 5x = 0$ is $x = y$,then the set of all values of $\cos y$ is

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