The general solution of 1+ sin^2 ⁡x=3 sin ⁡x cos ⁡x, tan ⁡x 12 is

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Accepted Answer

Correct answer: C. (c) : 1+ sin^2 ⁡x=3 sin ⁡x cos ⁡x ⇒ cos^2 ⁡x+ sin^2 ⁡x+ sin^2 ⁡x-3 sin ⁡x cos ⁡x=0 ⇒ cos^2 ⁡x+ sin^2 ⁡x-2 sin ⁡x cos ⁡x+ sin^2 ⁡x- sin ⁡x cos ⁡x=0 ⇒ ( sin ⁡x- cos ⁡x )^2+ sin ⁡x ( sin ⁡x- cos ⁡x )=0 ⇒ ( sin ⁡x- cos ⁡x ) ( sin ⁡x- cos ⁡x+ sin ⁡x )=0 ⇒ ( sin ⁡x- cos ⁡x ) (2 sin ⁡x- cos ⁡x )=0 ⇒ either sin ⁡x= cos ⁡x or sin ⁡x cos ⁡x= 12 ⇒ tan ⁡x=1 , tan ⁡x= 12 But tan ⁡x 12 (given) ⇒ tan ⁡x=1= tan ⁡ 4 ⇒x= nπ+ 4

$The general solution of 1 + sin^{2} x = 3 sin x \cdot cos x, tan x \neq = \frac{1}{2} is$

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