The number of solutions of the equation $\tan (x)+\sec (x)=2\cos (x)$ in the interval $[0,2\pi ]$ is

Solution

1. Rewrite the equation $\tan (x)+\sec (x)=2\cos (x)$ as $\frac{\sin (x)+1}{\cos (x)}=2\cos (x)$, noting that $\cos (x)\ne 0$.
2. Multiply by $\cos (x)$ to obtain $\sin (x)+1=2{\cos }^2(x)$, then substitute ${\cos }^2(x)=1-{\sin }^2(x)$ to get $2{\sin }^2(x)+\sin (x)-1=0$.
3. Factor the quadratic equation as $(2\sin (x)-1)(\sin (x)+1)=0$, which gives $\sin (x)=\frac{1}{2}$ or $\sin (x)=-1$.
4. For $\sin (x)=\frac{1}{2}$, we have $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$, both of which satisfy $\cos (x)\ne 0$.
5. For $\sin (x)=-1$, we have $x=\frac{3\pi }{2}$, but this makes $\cos (x)=0$, which is undefined for the original equation.
6. Hence the answer is (B).