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If $5 \cos 2 \theta+2 \cos ^2 \frac{\theta}{2}+1=0$, when $(0 < \theta < \pi)$, then the values of $\theta$ are :
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Verified Answer
The correct answer is:
$\frac{\pi}{3}, \pi-\cos ^{-1}\left(\frac{3}{5}\right)$
Hints : $5 \cos 2 \theta+1+\cos \theta+1=0$
$$
\begin{aligned}
& 5\left(2 \cos ^2 \theta-1\right)+\cos \theta+2=0 \\
& 10 \cos ^2 \theta+\cos \theta-3=0 \\
& (5 \cos \theta+3)(2 \cos \theta-1)=0 \\
& \cos \theta=\frac{1}{2} \\
& \theta=\frac{\pi}{3}
\end{aligned}
$$
$$
\mid \begin{aligned}
& \cos \theta=-\frac{3}{5} \\
& \theta=\cos ^{-1}\left(-\frac{3}{5}\right) \\
& =\pi-\cos ^{-1}\left(\frac{3}{5}\right)
\end{aligned}
$$
$$
\begin{aligned}
& 5\left(2 \cos ^2 \theta-1\right)+\cos \theta+2=0 \\
& 10 \cos ^2 \theta+\cos \theta-3=0 \\
& (5 \cos \theta+3)(2 \cos \theta-1)=0 \\
& \cos \theta=\frac{1}{2} \\
& \theta=\frac{\pi}{3}
\end{aligned}
$$
$$
\mid \begin{aligned}
& \cos \theta=-\frac{3}{5} \\
& \theta=\cos ^{-1}\left(-\frac{3}{5}\right) \\
& =\pi-\cos ^{-1}\left(\frac{3}{5}\right)
\end{aligned}
$$
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