Solve for x, $2 s i n^{3} x = c o s x$ .

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nπ, n ∈ I

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Solution

The correct option is A

Here $2 s i n^{2} x = \frac{c o s x}{s i n x}$
{Q sinx $\neq$ 0 because sin x = 0 does not satisfy the given equation}

or, $\frac{2}{c o s e c^{2} x} = c o t x o r 2 = c o t x (1 + c o t^{2} x)$
or, $c o t^{3} x + c o t x - 2 = 0$
By trial, cot x = 1 satisfies the equation.
$∴ (c o t x - 1) (c o t^{2} x + c o t x + 2) = 0$

$\Rightarrow c o t x = 1 o r c o t^{2} x + c o t x + 2 = 0$

When $c o t x = 1, t a n x = 1 \Rightarrow x = n π + \frac{π}{4}, n \in I .$

When $c o t^{2} x + c o t x + 2 = 0$ ,
cot x has no real value because D = 1- 8 < 0
$∴$ Solutions are $x = n π + \frac{π}{4}, n \in I$

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