Solve the following quadratic equations
Question:

If $0 \leq x<\frac{\pi}{2}$, then the number of values of $x$ for which $\sin x-\sin 2 x+\sin 3 x=0$, is

  1. 2

  2. 1

  3. 3

  4. 4


Correct Option: 1

Solution:

$\sin x-\sin 2 x+\sin 3 x=0$

$\Rightarrow(\sin x+\sin 3 x)-\sin 2 x=0$

$\Rightarrow 2 \sin x \cdot \cos x-\sin 2 x=0$

$\Rightarrow \sin 2 x(2 \cos x-1)=0$

$\Rightarrow \sin 2 x=0$ or $\cos x=\frac{1}{2}$

$\Rightarrow x=0, \frac{\pi}{3}$

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