Write the number of solutions of the equation tan

Given:

tanx + secx = 2 cosx

$\Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x$

$\Rightarrow \frac{\sin x+1}{\cos x}=2 \cos x$

$\Rightarrow \sin x+1=2 \cos ^{2} x$

$\Rightarrow \sin x=2 \cos ^{2} x-1$

$\Rightarrow 2\left(1-\sin ^{2} x\right)-1=\sin x$

$\Rightarrow 2-2 \sin ^{2} x-1=\sin x$

$\Rightarrow 1-2 \sin ^{2} x=\sin x$

$\Rightarrow 2 \sin ^{2} x+\sin x-1=0$

$\Rightarrow 2 \sin ^{2} x+2 \sin x-\sin x-1=0$

$\Rightarrow 2 \sin x(\sin x+1)-1(\sin x+1)=0$

$\Rightarrow(\sin x+1)(2 \sin x-1)=0$

$\Rightarrow \sin x+1=0$ or $2 \sin x-1=0$

$\Rightarrow \sin x=-1$ or $\sin x=\frac{1}{2}$

Now,

$\sin x=-1$

$\Rightarrow \sin x=\sin \left(\frac{3 \pi}{2}\right)$

$\Rightarrow x=n \pi+(-1)^{n} \frac{3 \pi}{2}, n \in Z$

Because it contains an odd multiple of $\frac{\pi}{2}$ and we know that tanx and secx are undefined on the odd multiple, this value will not satisfy the given equation.

And,

$\sin x=\frac{1}{2}$

$\Rightarrow \quad \sin x=\sin \left(\frac{\pi}{6}\right)$

$\Rightarrow \quad x=n \pi+(-1)^{n} \frac{\pi}{6}, n \in Z$

Now,

For $n=0, x=\frac{\pi}{6}$

For $n=1, x=\frac{11 \pi}{6}$

For other values of $n$, the condition is not true.

Hence, the given equation has two solutions in $[0,2 \pi]$.