# 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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