Question

A solution of the equation $5{\sin }^{2}x+3\sin x\cos x-3{\cos }^{2}x=2$ is-

• A. $2\pi +{\tan }^{-1}\frac{-3+\sqrt{69}}{6}$
• B. $7\pi +{\tan }^{-1}\frac{-3-\sqrt{69}}{6}$
• C. ${\tan }^{-1}\frac{-3+\sqrt{69}}{6}-\pi$
• D. ${\tan }^{-1}\frac{-3-\sqrt{69}}{6}-5\pi$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $2\pi +{\tan }^{-1}\frac{-3+\sqrt{69}}{6}$
B $7\pi +{\tan }^{-1}\frac{-3-\sqrt{69}}{6}$
C ${\tan }^{-1}\frac{-3-\sqrt{69}}{6}-5\pi$
D ${\tan }^{-1}\frac{-3+\sqrt{69}}{6}-\pi$

$5{\sin }^{2}x+3\sin x\cos x-3{\cos }^{2}x=2$
$5{\tan }^{2}x+3\tan x-3=2{\sec }^{2}x$ (Dividing by ${\cos }^{2}x$)
$3{\tan }^{2}x+3\tan x-5=0$
$\Rightarrow \tan x=\frac{-3\pm \sqrt{60}}{6}$
$\Rightarrow x={\tan }^{-1}\left(\frac{-3\pm \sqrt{60}}{6}\right)$
The general solution is
$x=n\pi +{\tan }^{-1}\left(\frac{-3\pm \sqrt{60}}{6}\right),n\in I$