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 = …
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