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

The given equation is equivalent to

          $7{\cos }^{2}x+\sin x\cos x-3\left({\sin }^{2}x+{\cos }^{2}x\right)=0$

or             $3{\sin }^{2}x-\sin x\cos x-4{\cos }^{2}x=0$

Since $\cos x = 0$ does not satisfy the given equation we divide throughout by ${\cos }^2 x$ and get

$\begin{array}{r}3{\tan }^{2}x-\tan x-4=0\\ \Rightarrow (3\tan x-4)(\tan x+1)=0\end{array}$

$\Rightarrow \tan x=4/3$ or $\tan x=-1$. That is $x=k\pi +{\tan }^{-1}(4/3)$

or $x=n\pi +3\pi /4(n,k\in \mathbf{I})$ . It should be noted that the solutions given by (b) or (c) are not complete;

 the general solution is given by ( d) only.