<s>[INST] 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?
 [/INST] 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 $x=b$
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