Question -12

On the interval $[-\frac{\pi}{4},\frac{\pi}{4}]$ the $\cos x$ is different from zero. So we can divide each row of the determinant by $\cos x$ and then multiply the whole determinant by $\cos^{3}x$ . Then the given equation becomes \cos^{3}x\left|\begin{array}{c,c,c}\tan x & 1 &1\\1 &\tan x&1\\1&1&\tan x\\\end{array}\right|=0. Dividing both sides of the equation by $\cos^{3}x$ - which is different from zero on $[-\frac{\pi}{4},\frac{\pi}{4}]$ - and finding the determinant we get the equation $\tan^{3}x- 3\tan x +2=0.$ Factoring $(\tan x -1)^2(\tan x+2)=0.$ Therefore $\tan x=1$ or $\tan x =-2$ . The second equation has not solutions on $[-\frac{\pi}{4},\frac{\pi}{4}]$ and the first has only one solution which is $x= \frac{\pi}{4}$ . Because of that the number of solutions of the original equation on $[-\frac{\pi}{4},\frac{\pi}{4}]$ is 1.

Solve determinant and divide by sinxcube. Put cotx equal to t and t does not lie in (-1,1). Factors of cubic are ( t-1 )square* (2 t+1)... -1/2 is rejected so ans is only 1...