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If the general solution of $\sin x+3 \sin 3 x+\sin 5 x=0$ is $x=y$ then the set of all values of $\cos y$ is
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Verified Answer
The correct answer is:
$\left\{-1,-\frac{1}{2}, \frac{1}{2}, 1\right\}$
$$
\begin{aligned}
& \text { } \sin x+\sin 5 x+3 \sin 3 x=0 \\
& \Rightarrow 2 \sin 3 x \cos 2 x+3 \sin 3 x=0 \\
& \Rightarrow \sin 3 x[+2 \cos 2 x+3]=0 \quad \Rightarrow \sin 3 x=0 \\
& \Rightarrow \quad x=0, \pi \\
& \text { So, } \quad y=0, \pi \Rightarrow \cos 0=1 \\
& \Rightarrow \quad \cos \pi=-1 \text { and } \cos 2 x=-\frac{3}{2} \\
& \Rightarrow \quad 2 \cos ^2 x-1=\left|-\frac{3}{2}\right| \Rightarrow 2 \cos ^2 x=1-\frac{3}{2} \\
& \Rightarrow \quad \cos ^2 x=\frac{1}{4} \quad \Rightarrow \quad \cos x= \pm \frac{1}{2} \\
& \Rightarrow \quad \cos y=1,-1, \frac{1}{2},-\frac{1}{2} \\
&
\end{aligned}
$$
\begin{aligned}
& \text { } \sin x+\sin 5 x+3 \sin 3 x=0 \\
& \Rightarrow 2 \sin 3 x \cos 2 x+3 \sin 3 x=0 \\
& \Rightarrow \sin 3 x[+2 \cos 2 x+3]=0 \quad \Rightarrow \sin 3 x=0 \\
& \Rightarrow \quad x=0, \pi \\
& \text { So, } \quad y=0, \pi \Rightarrow \cos 0=1 \\
& \Rightarrow \quad \cos \pi=-1 \text { and } \cos 2 x=-\frac{3}{2} \\
& \Rightarrow \quad 2 \cos ^2 x-1=\left|-\frac{3}{2}\right| \Rightarrow 2 \cos ^2 x=1-\frac{3}{2} \\
& \Rightarrow \quad \cos ^2 x=\frac{1}{4} \quad \Rightarrow \quad \cos x= \pm \frac{1}{2} \\
& \Rightarrow \quad \cos y=1,-1, \frac{1}{2},-\frac{1}{2} \\
&
\end{aligned}
$$
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