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=2−2sin2x ⇒2sin2x+sinx−1=0 ⇒(2sinx−1)(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.