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In \(\triangle P Q R\), let \(\angle P>\angle Q\). If the radian measures of \(\angle P\) and \(\angle Q\) satisfy the equation \(4 \sin ^3 x-3 \sin x+a=0,0 < a < 1\), then the radian measure of \(\angle R\) is
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Verified Answer
The correct answer is:
\(\frac{2 \pi}{3}\)
Given, \(4 \sin ^3 x-3 \sin x+a=0\)
\(\begin{aligned}
\Rightarrow & & a & =3 \sin x-4 \sin ^3 x \\
\Rightarrow & & a & =\sin 3 x \\
& \ddots & \sin 3 P & =a=\sin 3 Q
\end{aligned}\)
[since \(P\) and \(Q\) satisfy the equation]
\(\begin{array}{lc}
\Rightarrow & \sin 3 P=\sin 3 Q \\
\Rightarrow & \sin 3 P=\sin (\pi-3 Q) \\
\Rightarrow & 3 P=\pi-3 Q \\
\Rightarrow & 3(P+Q)=\pi \\
\Rightarrow & P+Q=\frac{\pi}{3} \quad \ldots (i) \\
\because & \angle P+\angle Q+\angle R=\pi \\
\therefore & \angle R=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}
\end{array}\)
\(\begin{aligned}
\Rightarrow & & a & =3 \sin x-4 \sin ^3 x \\
\Rightarrow & & a & =\sin 3 x \\
& \ddots & \sin 3 P & =a=\sin 3 Q
\end{aligned}\)
[since \(P\) and \(Q\) satisfy the equation]
\(\begin{array}{lc}
\Rightarrow & \sin 3 P=\sin 3 Q \\
\Rightarrow & \sin 3 P=\sin (\pi-3 Q) \\
\Rightarrow & 3 P=\pi-3 Q \\
\Rightarrow & 3(P+Q)=\pi \\
\Rightarrow & P+Q=\frac{\pi}{3} \quad \ldots (i) \\
\because & \angle P+\angle Q+\angle R=\pi \\
\therefore & \angle R=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}
\end{array}\)
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