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?
 A: Following on from $$(\sin x +1)(2\sin x-1)=0$$
Note it's useful to sketch the graph of $\sin x$ in the given interval:

We get that either:
$\sin x=-1\implies x=\frac{3\pi}{2},\quad$or
$\sin x =\frac{1}{2}\implies x=\frac{\pi}{6}\quad\text{or}\quad x=\frac{5\pi}{6}$
Now we must be careful since we had $\cos x$ in the denominator in the original equation, we require that $\cos x \neq 0$. This eliminates the solution $x=\frac{3\pi}{2}$. Therefore the total number of solutions is $2$ - so the answer is $(c)$.
A: Obviously $x$ has possible values of $\frac{3 \pi}{2}$ (by $\sin x=-1$) and $\frac{ \pi}{6}$ , $\frac{5 \pi}{6}$ by $\sin x=\frac{1}{2}$
Now put These roots in intial conditions to verify if they satisfy it or not.
Note: $(x-a)(x-b)=0$ means $x=a$ or $x=b$
A: 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
