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

Question

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

Aicharya · Accepted Answer

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] $\begin{array}{lc} \Rightarrow & \sin 3 P=\sin 3 Q \ \Rightarrow & \sin 3 P=\sin (\pi-3 Q) \ \Rightarrow & 3 P=\pi-3 Q \ \Rightarrow & 3(P+Q)=\pi \ \Rightarrow & P+Q=\frac{\pi}{3} \quad \ldots (i) \ \because & \angle P+\angle Q+\angle R=\pi \ \therefore & \angle R=\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \end{array}$

In \(\triangle P Q R\), let \(\angle P>\angle Q\). If the radian measures of \(\angle P\) and \(\angle Q\)…

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