Find the number of solutions of tanx + secx = 2cosx in [0, \(2\pi\)].

Solution :

Here, tanx + secx = 2cosx \(\implies\) sinx + 1 = \(2cos^2x\)

\(\implies\) \(2sin^2x\) + sinx – 1 = 0 \(\implies\) sinx = \(1\over 2\), -1

But sinx = -1 \(\implies\) x = \(3\pi\over 2\) for which tanx + secx = 2cosx is not defined.

Thus, sinx = \(1\over 2\) \(\implies\) x = \(\pi\over 6\), \(5\pi\over 6\)

\(\implies\) number of solutions of tanx + secx = 2cosx is 2.