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If the general solution of $\sin 5 x=\cos 2 x$ is of the form $a_n \cdot \frac{\pi}{2}$ for $n=0, \pm 1, \pm 2, \ldots .$. , then $a_n=$
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Verified Answer
The correct answer is:
$\frac{2 n+(-1)^n}{5+2(-1)^n}$
Given,
$$
\Rightarrow \quad \begin{aligned}
& \sin 5 x=\cos 2 x \\
& \quad \sin 5 x=\sin \left(\frac{\pi}{2}-2 x\right)
\end{aligned}
$$
$$
\begin{array}{lrl}
\Rightarrow & 5 x=n \pi+(-1)^n\left(\frac{\pi}{2}-2 x\right) \\
\Rightarrow & 5 x=n \pi+(-1)^n \frac{\pi}{2}-(-1)^n 2 x \\
\Rightarrow & 5 x+(-1)^n(2 x)=\frac{\pi}{2}\left\{2 n+(-1)^n\right\} \\
\Rightarrow & x\left(5+(-1)^n 2=\frac{\pi}{2}\left(2 n+(-1)^n\right)\right. \\
\Rightarrow & x=\frac{\pi}{2}\left(\frac{2 n+(-1)^n}{5+2(-1)^n}\right) \Rightarrow x=a_n \cdot \frac{\pi}{2}
\end{array}
$$
So,
$$
a_n=\frac{2 n+(-1)^n}{5+2(-1)^n}
$$
$$
\Rightarrow \quad \begin{aligned}
& \sin 5 x=\cos 2 x \\
& \quad \sin 5 x=\sin \left(\frac{\pi}{2}-2 x\right)
\end{aligned}
$$
$$
\begin{array}{lrl}
\Rightarrow & 5 x=n \pi+(-1)^n\left(\frac{\pi}{2}-2 x\right) \\
\Rightarrow & 5 x=n \pi+(-1)^n \frac{\pi}{2}-(-1)^n 2 x \\
\Rightarrow & 5 x+(-1)^n(2 x)=\frac{\pi}{2}\left\{2 n+(-1)^n\right\} \\
\Rightarrow & x\left(5+(-1)^n 2=\frac{\pi}{2}\left(2 n+(-1)^n\right)\right. \\
\Rightarrow & x=\frac{\pi}{2}\left(\frac{2 n+(-1)^n}{5+2(-1)^n}\right) \Rightarrow x=a_n \cdot \frac{\pi}{2}
\end{array}
$$
So,
$$
a_n=\frac{2 n+(-1)^n}{5+2(-1)^n}
$$
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