MHT CET · Maths · Trigonometric Equations
The number of possible solutions of \(\sin \theta+\sin 4 \theta+\sin 7 \theta=0, \theta \in(0, \pi)\) are
- A 3
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \sin 7 \theta+\sin \theta+\sin 4 \theta=0 \\
& \Rightarrow 2 \sin 4 \theta \cos 3 \theta+\sin 4 \theta=0 \\
& \Rightarrow \sin 4 \theta(2 \cos 3 \theta+1)=0 \\
& \Rightarrow \sin 4 \theta=0 \quad \text { or } \cos 3 \theta=\frac{-1}{2} \\
& \Rightarrow \sin 4 \theta=0 \quad \text { or } \cos 3 \theta=\cos \left(\frac{2 \pi}{3}\right) \\
& \Rightarrow 4 \theta=\mathrm{n} \pi \quad \text { or } 3 \theta=2 \mathrm{n} \pi \pm \frac{2 \pi}{3} \\
& \Rightarrow \theta=\frac{\mathrm{n} \pi}{4} \quad \text { or } \quad \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{2 \pi}{9} \\
& \quad \theta=\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{4 \pi}{9}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{8 \pi}{9} \ldots[\because \theta \in(0, \pi)]
\end{aligned}\)
\(\therefore \quad\) Number of solutions \(=6\)
& \sin 7 \theta+\sin \theta+\sin 4 \theta=0 \\
& \Rightarrow 2 \sin 4 \theta \cos 3 \theta+\sin 4 \theta=0 \\
& \Rightarrow \sin 4 \theta(2 \cos 3 \theta+1)=0 \\
& \Rightarrow \sin 4 \theta=0 \quad \text { or } \cos 3 \theta=\frac{-1}{2} \\
& \Rightarrow \sin 4 \theta=0 \quad \text { or } \cos 3 \theta=\cos \left(\frac{2 \pi}{3}\right) \\
& \Rightarrow 4 \theta=\mathrm{n} \pi \quad \text { or } 3 \theta=2 \mathrm{n} \pi \pm \frac{2 \pi}{3} \\
& \Rightarrow \theta=\frac{\mathrm{n} \pi}{4} \quad \text { or } \quad \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{2 \pi}{9} \\
& \quad \theta=\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{4 \pi}{9}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{8 \pi}{9} \ldots[\because \theta \in(0, \pi)]
\end{aligned}\)
\(\therefore \quad\) Number of solutions \(=6\)
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