MHT CET · Maths · Trigonometric Equations
The number of solutions of \(\sin x+\sin 3 x+\sin 5 x=0\) in the interval \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\) is
- A
- B
- C
- D
Answer & Solution
Correct Answer
(B)
Step-by-step Solution
Detailed explanation
\((\sin x +\sin 5 x )+\sin 3 x =0 \)
\( 2 \sin 3 x \cos 2 x +\sin 3 x =0 \)
\( \sin 3 x (2 \cos 2 x +1)=0 \)
\( \sin 3 x =0 \text { and } 2 \cos 2 x +1=0 \)
\( \sin 3 x=0 \)
Graph \(f ( x )=\sin 3 x\)
\(x=\frac{2 \pi}{3}\) or \(x=\pi\) or \(x=\frac{4 \pi}{3}\),
\(\therefore \sin 3 x=0\) have 3 solutions
Graph \(f(x)=\cos 2 x\)
\(\cos 2 x=-\frac{1}{2}\)
\(\therefore x=\frac{2 \pi}{3}\) and \(\frac{4 \pi}{3}\)
So the solution of given equation in \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\) are \(\frac{2 \pi}{3}\) and \(\frac{4 \pi}{3}\)
\( 2 \sin 3 x \cos 2 x +\sin 3 x =0 \)
\( \sin 3 x (2 \cos 2 x +1)=0 \)
\( \sin 3 x =0 \text { and } 2 \cos 2 x +1=0 \)
\( \sin 3 x=0 \)
Graph \(f ( x )=\sin 3 x\)
\(x=\frac{2 \pi}{3}\) or \(x=\pi\) or \(x=\frac{4 \pi}{3}\),
\(\therefore \sin 3 x=0\) have 3 solutions
Graph \(f(x)=\cos 2 x\)
\(\cos 2 x=-\frac{1}{2}\)
\(\therefore x=\frac{2 \pi}{3}\) and \(\frac{4 \pi}{3}\)
So the solution of given equation in \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\) are \(\frac{2 \pi}{3}\) and \(\frac{4 \pi}{3}\)
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