The number of solutions of the equation tanx+secx=2cosx in the interval [0,2π] is
A
1
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B
2
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C
3
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D
4
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Solution
The correct option is B2 tanx+secx=2cosx ⇒sinxcosx+1cosx=2cosx ⇒sinx+1=2cos2x for cosx≠0 ⇒sinx+1=2−2sin2x⇒2sin2x+sinx−1=0⇒(2sinx−1)(sinx+1=0) ⇒sinx=12 or sinx=−1 (not possible as then cosx=0 and secx,tanx are not defined.) x=π6,5π6 Hence, number of solutions =2