If $sin (\frac{3 x}{2}) + cos x = 2$ then number of solution of this equation in $[- π, π]$ is

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Solution

The correct option is A 0
$\begin{matrix} sin (\frac{3 x}{2}) + cos x = 2 f o r s o l^{n} : sin (\frac{3 x}{2}) = 1 a n d cos x = 1 \frac{3 x}{2} = (2 m π + \frac{π}{2}) a n d x = 2 n π [n = - 1, 0, 1] ∴ x = - 2 π, 0, 2 π F o r x = 0 sin (\frac{3 x}{2}) = 0 ∴ N o s o l u t i o n i n [- π, π] H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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