The general solution of 1+sin ^{2} x=3 sin x | vedclass

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Question

(D) Given equation: $1+\sin ^{2} x=3 \sin x \cdot \cos x$,with $	an x 
eq \frac{1}{2}$.Dividing both sides by $\cos ^{2} x$ (assuming $\cos x 
eq 0

Vedclass · Accepted Answer

$n \pi+\frac{\pi}{4}, n \in Z$

Vedclass · Answer

(D) Given equation: $1+\sin ^{2} x=3 \sin x \cdot \cos x$,with $	an x 
eq \frac{1}{2}$.Dividing both sides by $\cos ^{2} x$ (assuming $\cos x 
eq 0$):$\frac{1}{\cos ^{2} x} + \frac{\sin ^{2} x}{\cos ^{2} x} = 3 \frac{\sin x \cdot \cos x}{\cos ^{2} x}$$\sec ^{2} x + 	an ^{2} x = 3 	an x$Since $\sec ^{2} x = 1 + 	an ^{2} x$,we have:$1 + 	an ^{2} x + 	an ^{2} x = 3 	an x$$2 	an ^{2} x - 3 	an x + 1 = 0$Factoring the quadratic equation:$2 	an ^{2} x - 2 	an x - 	an x + 1 = 0$$2 	an x(	an x - 1) - 1(	an x - 1) = 0$$(	an x - 1)(2 	an x - 1) = 0$So,$	an x = 1$ or $	an x = \frac{1}{2}$.Given the condition $	an x 
eq \frac{1}{2}$,we must have $	an x = 1$.The general solution for $	an x = 1$ is $x = n \pi + \frac{\pi}{4}$,where $n \in Z$.

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

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