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

Equation: \(\tan x+\sec x=2 \cos x\), in the interval \([0,2 \pi]\).
Step 1: Substitute \(\sec x=\frac{1}{\cos x}\) and \(\tan x=\frac{\sin x}{\cos x}\) :
\(\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \)
\( \frac{\sin x+1}{\cos x}=2 \cos x\)
Step 2: Multiply through by \(\cos x\) (assuming \(\cos x \neq 0\) ):
\(\sin x+1=2 \cos ^2 x\)
Step 3: Use \(\cos ^2 x=1-\sin ^2 x\) :
\(\sin x+1=2\left(1-\sin ^2 x\right) \)
\( \sin x+1=2-2 \sin ^2 x\)
Step 4: Rearrange into a quadratic in \(\sin x\) :
\(2 \sin ^2 x+\sin x-1=0\)
Step 5: Solve the quadratic equation:
\(\sin x=\frac{-1 \pm \sqrt{1-4(2)(-1)}}{2(2)} \)
\( \sin x=\frac{-1 \pm \sqrt{9}}{4} \)
\( \sin x=\frac{-1+3}{4} \text { or } \sin x=\frac{-1-3}{4} \)
\( \sin x=\frac{1}{2} \text { or } \sin x=-1\)
Step 6: Find solutions in \([0,2 \pi]\) :
- For \(\sin x=\frac{1}{2}, x=\frac{\pi}{6}, \frac{5 \pi}{6}\).
- For \(\sin x=-1, x=\frac{3 \pi}{2}\).
Answer: \(\left\{\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}\right\}\), Option 3.