JEE Mains · Maths · STD 11 - Trigonometrical equations
The sum of all values of \(x\) in \([0,2 \pi]\), for which \(\sin x+\sin 2 x+\sin 3 x+\sin 4 x=0\), is equal to:
- A \(11 \pi\)
- B \(12 \pi\)
- C \(8 \pi\)
- D \(9 \pi\)
Answer & Solution
Correct Answer
(D) \(9 \pi\)
Step-by-step Solution
Detailed explanation
\((\sin x+\sin 4 x)+(\sin 2 x+\sin 3 x)=0\) \(\Rightarrow 2 \sin \frac{5 x}{2}\left\{\cos \frac{3 x}{2}+\cos \frac{x}{2}\right\}=0\) \(\Rightarrow 2 \sin \frac{5 x}{2}\left\{2 \cos x \cos \frac{x}{2}\right\}=0\)…
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