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Question

The number of solutions of the equation tanx+secx=2cosx lying in the interval [0,2π] is

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Solution

The given equation is tanx+secx=2cosx
sinxcosx+1cosx=2cosx
sinx+1=2cos2x=22sin2x
2sin2x+sinx1=0
(2sinx1)(sinx+1)=0
sinx=1,12
x=π6,5π6,3π2[0,2π]
But for x=3π/2, tanx and secx are not defined.
Therefore, there are only two solutions.

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