If 4 cos 2 x sin x-2 sin 2 x-3 sin x- then x-n - Z

The correct option is A 4 cos^2 x sin x 2 sin^2 x=3 sin x ⇒ sin x(4 cos^2 x 2 sin x 3)=0 ⇒ sin x=0 or 4 sin^2 x+2 sin x 1=0 ⇒ sin x=0 or 4 sin^2 x+2 sin x 1=0 ...

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Byju's Answer

Standard XI

Mathematics

General Solution of tan theta = tan alpha

If 4 cos 2 x ...

Question

$I f 4 c o s^{2} x s i n x - 2 s i n^{2} x = 3 s i n x, t h e n x = (n ϵ Z)$

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Solution

The correct option is A
$4 c o s^{2} x s i n x - 2 s i n^{2} x = 3 s i n x \Rightarrow s i n x (4 c o s^{2} x - 2 s i n x - 3) = 0 \Rightarrow s i n x = 0 o r 4 s i n^{2} x + 2 s i n x - 1 = 0 \Rightarrow s i n x = 0 o r 4 s i n^{2} x + 2 s i n x - 1 = 0 \Rightarrow s i n x = \frac{- 2 \pm \sqrt{4 + 16}}{8} = \frac{- 1 \pm \sqrt{5}}{4} = \frac{\sqrt{5} - 1}{4} \Rightarrow x = n π + (- 1)^{n} \frac{π}{10^{'}} n ϵ Z .$

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If 4 cos2x sin x−2 sin2x=3 sin x,then x=(nϵZ)

The correct option is A 4 cos2x sin x−2 sin2x=3 sin x⇒sin x(4 cos2x−2 sin x−3)=0⇒sin x=0 or 4 sin2x+2 sin x−1=0⇒sin x=0 or 4 sin2x+2 sin x−1=0⇒sin x=−2±√4+168=−1±√54=√5−14⇒x=nπ+(−1)nπ10′nϵZ.

$I f 4 c o s^{2} x s i n x - 2 s i n^{2} x = 3 s i n x, t h e n x = (n ϵ Z)$