MHT CET · Maths · Trigonometric Equations
The solutions of \(\sin x+\sin 5 x=\sin 3 x\) in \(\left(0, \frac{\pi}{2}\right)\) are
- A \(\frac{\pi}{4}, \frac{\pi}{10}\)
- B \(\frac{\pi}{6}, \frac{\pi}{3}\)
- C \(\frac{\pi}{4}, \frac{\pi}{12}\)
- D \(\frac{\pi}{8}, \frac{\pi}{16}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{6}, \frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin x+\sin 5 x=\sin 3 x \\ & \Rightarrow 2 \sin 3 x \cos 2 x=\sin 3 x \\ & \Rightarrow \sin 3 x(2 \cos 2 x-1)=0 \\ & \Rightarrow \sin 3 x=0 \text { or } \cos 2 x=\frac{1}{2}=\cos \frac{\pi}{3} \\ & \Rightarrow 3 x=\mathrm{n} \pi \text { or } 2 x=2 \mathrm{n} \pi \pm \frac{\pi}{3} \\ & \Rightarrow x=\frac{\mathrm{n} \pi}{3} \text { or } x=\mathrm{n} \pi \pm \frac{\pi}{6}\end{aligned}\)
\(\Rightarrow x=\frac{\pi}{3}, \frac{\pi}{6} \quad \quad \ldots\left[\because x \in\left(0, \frac{\pi}{2}\right)\right]\)
\(\Rightarrow x=\frac{\pi}{3}, \frac{\pi}{6} \quad \quad \ldots\left[\because x \in\left(0, \frac{\pi}{2}\right)\right]\)
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