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.


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