Question

The complete solution of the equation $7{\cos }^{2}x+\sin x\cos x-3=0$ is given by

• A. $n\pi +\frac{\pi }{2}\left(n\epsilon I\right)$
• B. $n\pi -\frac{\pi }{4}\left(n\epsilon I\right)$
• C. $n\pi +ta{n}^{-1}\left(\frac{4}{3}\right)\left(n\epsilon I\right)$
• D. $n\pi +\frac{3\pi }{4},k\pi +ta{n}^{-1}\left(\frac{4}{3}\right)(n,k\epsilon I)$

Answer 1

The correct option is D $n\pi +\frac{3\pi }{4},k\pi +ta{n}^{-1}\left(\frac{4}{3}\right)(n,k\epsilon I)$

The given equation is equivalent to
$7{\cos }^{2}x+\sin x\cos x-3({\sin }^{2}x+{\cos }^{2}x)=0$
or $3{\sin }^{2}x-\sin x\cos x-4{\cos }^{2}x=0$
Since $\cos x=0$ does not satisfy the given equation .
Divide throughout by $co{s}^{2}x$ we get
$3{\tan }^{2}x-\tan x-4=0$
$\Rightarrow (3\tan x-4)(\tan x+1)=0$
$\Rightarrow \tan x=\frac{4}{3}$ or $\tan x=-1$
$x=k\pi +ta{n}^{-1}\frac{4}{3}$
or $x=n\pi +\frac{3\pi }{4}(n,k\in I)$
Hence, the general solution is given by (d) only.