The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is: The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is:
$a$. $0$
$b$. $1$
$c$. $2$
$d$. $3$
My Attempt:
$$\tan x +\sec x=2\cos x$$
$$\dfrac {\sin x}{\cos x}+\dfrac {1}{\cos x}=2\cos x$$
$$\sin x + 1=2\cos^2 x$$
$$\sin x +1=2-2\sin^2 x$$
$$2\sin^2 x +\sin x - 1=0$$
$$2\sin^2 x +2\sin x -\sin x - 1=0$$
$$2\sin x (\sin x +1) -1(\sin x+1)=0$$
$$(\sin x +1)(2\sin x-1)=0$$
So, what's the next?

The answer must be 2 
Since tanx and secx domain would consider every x, rejecting (2n+1)π/2 values in the interval [0,2π] such values would be rejected from the solution.
In the last step sinx=1 the solution would be 3π/2 and this would be rejected due to domain condition. The other part, that is, sinx=1/2 has solution x=5π/6 and x= π/6 in the Interval [0,2π]. 
Thus the number of solutions are 2
