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Trigonometrical Equations
hard
$[0,2 \pi]$ में $x$ के सभी मानों, जिनके लिए $\sin x +\sin 2 x +\sin 3 x +\sin 4 x =0$ है, का योग है
A
$11 \pi$
B
$12 \pi$
C
$8 \pi$
D
$9 \pi$
(JEE MAIN-2021)
Solution
$(\sin x+\sin 4 x)+(\sin 2 x+\sin 3 x)=0$
$\Rightarrow 2 \sin \frac{5 x}{2}\left\{\cos \frac{3 x}{2}+\cos \frac{x}{2}\right\}=0$
$\Rightarrow 2 \sin \frac{5 x}{2}\left\{2 \cos x \cos \frac{x}{2}\right\}=0$
$2 \sin \frac{5 x}{2}=0 \Rightarrow \frac{5 x}{2}=0, p, 2 \pi, 3 \pi, 4 \pi, 5 \pi$
$\Rightarrow x=0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \frac{6 \pi}{5}, \frac{8 \pi}{5}, 2 \pi$
$\cos \frac{x}{2}=0 \Rightarrow \frac{x}{2}=\frac{x}{2} \Rightarrow x=\pi$
$\cos x=0 \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2}$
So cum $=6 \pi+\pi+2 \pi=9 \pi$
Standard 11
Mathematics