Q68 / 180 · KCET 2010 · Mathematics
The general solution of $1+\sin ^{2} x=3 \sin x \cdot \cos x, \tan x \neq \frac{1}{2}$, is
- A$2 \mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$
- B$2 \mathrm{n} \pi-\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$
- C$\mathrm{n} \pi-\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$
- Correct answerD$\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$
Solution
$$
1+\sin ^{2} x=3 \sin x \cdot \cos x, \tan x \neq \frac{1}{2}
$$
Divided by $\cos ^{2} \mathrm{x}$ on both sides,
$$
\begin{gathered}
\frac{1}{\cos ^{2} \mathrm{x}}+\frac{\sin ^{2} \mathrm{x}}{\cos ^{2} \mathrm{x}}=3 \frac{\sin \mathrm{x} \cdot \cos \mathrm{x}}{\cos \mathrm{x} \cdot \cos \mathrm{x}} \\
\sec ^{2} \mathrm{x}+\tan ^{2} \mathrm{x}=3 \tan \mathrm{x} \\
1+\tan ^{2} \mathrm{x}+\tan ^{2} \mathrm{x}=3 \tan \mathrm{x} \\
2 \tan ^{2} \mathrm{x}-3 \tan \mathrm{x}+1=0 \\
2 \tan ^{2} \mathrm{x}-2 \tan \mathrm{x}-\tan \mathrm{x}+1=0 \\
2 \tan \mathrm{x}(\tan \mathrm{x}-1)-1(\tan \mathrm{x}-1)=0 \\
(\tan \mathrm{x}-1)(2 \tan \mathrm{x}-1)=0 \\
\tan \mathrm{x}=1, \frac{1}{2} \\
\mathrm{We take,} \quad \tan \mathrm{x}=1 \\
\left(\because \tan \mathrm{x} \neq \frac{1}{2}\right) \\
\mathrm{x}=\mathrm{tan}(\pi / 4)
\end{gathered}
$$
$\mathrm{x}=\mathrm{n} \pi+\pi / 4, \mathrm{n} \in \mathrm{Z}$
Hence, the correct answer is option: 'D'.