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Q. The solution of the equation cos2x2cosx=4sinx+4sin2x(0xπ) is

Inverse Trigonometric Functions

Solution:

We have, cos2x2cosx=4sinx+4sin2x
cos2x4sin2x=4sinx+2cosx
(cosx+2sinx)(cosx2sinx2)=0
tanx=12,cosx2sinx=2
1tan2x21+tan2x222tanx21+tan2x2=2
1tan2x24tanx2=2+2tan2x23tan2x2+4tanx2+1=0
(3tanx2+1)(tanx2+1)=0tanx2=13 or 1
x2=tan1(13) or π4x=2tan1(13),π2]