Question

Solve $2{\sin }^{2}x-5\sin x\cos x-8{\cos }^{2}x=-2$

• A. $x=m\pi +{\tan }^{-1}(-\frac{3}{2})$, where $mϵZ$
• B. $x=m\pi +{\tan }^{-1}(-\frac{3}{5})$, where $mϵZ$
• C. $x=m\pi +{\tan }^{-1}(-\frac{3}{7})$, where $mϵZ$
• D. $x=m\pi +{\tan }^{-1}(-\frac{3}{4})$, where $mϵZ$

Answer 1

Answer: (D)

$x=m\pi +{\tan }^{-1}(-\frac{3}{4})$, where $mϵZ$

If $\cos x=0$, then

$2{\sin }^{2}x=-2$ or ${\sin }^{2}x=-1$, which is not possible

Clearly, $\cos x\ne 0$

Given, $2{\sin }^{2}x-5\sin x\cos x-8{\cos }^{2}x=-2$

Dividing both sides by ${\cos }^{2}x$, we get

$2{\tan }^{2}x-5\tan x-8=2{\sec }^{2}x$

$2{\tan }^{2}x-5\tan x-8+2(1+{\tan }^{2}x)=0$

$4{\tan }^{2}x-5\tan x-6=0$

$(\tan x-2)(4\tan x+3)=0$

Now, $\tan x-2=0$

Let $\tan x=2=\tan \alpha$

$\Rightarrow$ $x=n\pi +\alpha =n\pi {\tan }^{-1}2$

$4\tan x+3=0$ or $\tan x=-\frac{3}{4}=\tan \beta$(let)

$\Rightarrow$ $x=n\pi +\beta =m\pi +{\tan }^{-1}(-\frac{3}{4})$, where $n,mϵZ$