Question

Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].

Answer 1

Given:
tanx + secx = 2 cosx
\[\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x\]
\[ \Rightarrow \frac{\sin x + 1}{\cos x} = 2 \cos x\]
\[ \Rightarrow \sin x + 1 = 2 \cos^2 x\]
\[ \Rightarrow \sin x = 2 \cos^2 x - 1\]

\[\Rightarrow 2\left( 1 - \sin^2 x \right) - 1 = \sin x\]

\[ \Rightarrow 2 - 2 \sin^2 x - 1 = \sin x\]

\[ \Rightarrow 1 - 2 \sin^2 x = \sin x\]

\[ \Rightarrow 2 \sin^2 x + \sin x - 1 = 0\]

\[ \Rightarrow 2 \sin^2 x + 2\sin x - \sin x - 1 = 0\]

\[ \Rightarrow 2\sin x\left( \sin x + 1 \right) - 1\left( \sin x + 1 \right) = 0\]

\[ \Rightarrow \left( \sin x + 1 \right)\left( 2\sin x - 1 \right) = 0\]

\[ \Rightarrow \sin x + 1 = 0\text{ or }2\sin x - 1 = 0\]

\[ \Rightarrow \sin x = - 1\text{ or }\sin x = \frac{1}{2}\]
Now, 
\[\sin x = - 1\]
\[ \Rightarrow \sin x = \sin\left( \frac{3\pi}{2} \right)\]
\[ \Rightarrow x = n\pi + \left( - 1 \right)^n \frac{3\pi}{2}, n \in Z\]
Because it contains an odd multiple of `pi/2` and we know that tan x and sec x are undefined on the odd multiple, this value will not satisfy the given equation.
And,

\[\sin x = \frac{1}{2}\]

\[ \Rightarrow \sin x = \sin\left( \frac{\pi}{6} \right)\]

\[ \Rightarrow x = n\pi + \left( - 1 \right)^n \frac{\pi}{6}, n \in Z\]

Now, 

\[\text{ For } n = 0, x = \frac{\pi}{6}\]

\[\text{ For }n = 1, x = \frac{11\pi}{6} \]

For other values of n, the condition is not true.
Hence, the given equation has two solutions in 

\[\left[ 0, 2\pi \right]\]