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If the possible solution of the equation $2 \cos ^2 x+3 \sin x-3=0$ constitute two unequal angles of a triangle, then the third angle of that triangle is
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Verified Answer
The correct answer is:
$\frac{\pi}{3}$
Given,
$\begin{gathered}
2 \cos ^2 x+3 \sin x-3=0 \Rightarrow 2-2 \sin ^2 x+3 \sin x-3=0 \\
2 \sin ^2 x-3 \sin x+1=0 \Rightarrow \quad(2 \sin x-1)(\sin x-1)=0 \\
\sin x=\frac{1}{2} \text { and } \sin x=1 \quad \Rightarrow \quad x=\frac{\pi}{6} \text { and } x=\frac{\pi}{2}
\end{gathered}$
$\frac{\pi}{6}, \frac{\pi}{2}$ are angle of triangle
$\therefore \text { Third angle }=\pi-\left(\frac{\pi}{6}+\frac{\pi}{2}\right)=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}$
$\begin{gathered}
2 \cos ^2 x+3 \sin x-3=0 \Rightarrow 2-2 \sin ^2 x+3 \sin x-3=0 \\
2 \sin ^2 x-3 \sin x+1=0 \Rightarrow \quad(2 \sin x-1)(\sin x-1)=0 \\
\sin x=\frac{1}{2} \text { and } \sin x=1 \quad \Rightarrow \quad x=\frac{\pi}{6} \text { and } x=\frac{\pi}{2}
\end{gathered}$
$\frac{\pi}{6}, \frac{\pi}{2}$ are angle of triangle
$\therefore \text { Third angle }=\pi-\left(\frac{\pi}{6}+\frac{\pi}{2}\right)=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}$
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