Question

The general solution of $1+\operatorname{Sin}^{2} x=3 \operatorname{Sin} x \cdot \operatorname{Cos} x, \operatorname{Tan} x \neq \frac{1}{2}$ is ......

• A. $n\pi-\frac{\pi}{4},n\in Z$
• B. $n\pi+\frac{\pi}{4},n\in Z$
• C. $2n\pi+\frac{\pi}{4},n\in Z$
• D. $2n\pi-\frac{\pi}{4},n\in Z$

Answer 1

The Correct Option is B

$1+\sin ^{2} x=3 \sin x \cdot \cos x, \tan x \neq \frac{1}{2}$ Divided by $\cos ^{2} x$ on both sides, $\frac{1}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}=3 \frac{\sin x \cdot \cos x}{\cos x \cdot \cos x}$ $\sec ^{2} x+\tan ^{2} x=3 \tan x$ $1+\tan ^{2} x+\tan ^{2} x=3 \tan x$ $2 \tan ^{2} x-3 \tan x+1=0$ $2 \tan ^{2} x-2 \tan x-\tan x+1=0$ $2 \tan x(\tan x-1)-1(\tan x-1)=0$ $(\tan x-1)(2 \tan x-1)=0$ $\tan x=1, \frac{1}{2}$ We take, $\tan x =1 \left(\because \tan x \neq \frac{1}{2}\right)$ $\tan x =\tan (\pi / 4)$ $x =n \pi+\pi / 4, n \in Z$