If $s i n^{2} 4 x + c o s^{2} x = 2 s i n 4 x . c o s^{4} x$ then number of values of x satisfying, if $x \in [- 2 π, 2 π]$ is

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Solution

The correct option is A
0

$s i n^{2} 4 x + c o s^{2} x = 2 s i n 4 x c o s^{4} x \Rightarrow s i n^{2} 4 x - 2 s i n 4 x c o s^{4} x + c o s^{2} x ∴ s i n 4 x = \frac{2 c o s^{4} x \pm \sqrt{4 c o s^{8} x - 4 c o s^{2} x}}{2} s i n 4 x = c o s^{4} x \pm \sqrt{c o s^{2} x (c o s^{6} x - 1)} ∴ c o s^{6} x = 1 s i n 4 x = 1 \Rightarrow 4 x = (4 n_{2} + 1) \frac{π}{2} ∴ c o s x = \pm 1 x = (4 n_{2} + 1) \frac{π}{8} ∴ x = n_{1} π, n, \in Z ∴ n_{1} π = (4 n_{2} + 1) \frac{π}{8}$
Not possible,
$∴$ No solution

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