The number of solution of 3^(sin2x +2 cos ^2x) +3^1-sin2x+2sin^2x=28 in [0,2pi] is

Answers (1)

Solution:    We have ,

                   3^sin 2x+2cos^2x+3^1-sin 2x+2sin^2x=28

          Let    t=sin^2x+2cos^2x.  Then   3^t+frac273^t=28

       	herefore        sin^2 x+2cos ^2x=0,       sin 2x+ cos 2x =-1

        Rightarrow           cos (2x- fracpi4)=cos frac3pi4;       2x-fracpi4=2n pipm frac3pi4

          	herefore                                          x=fracpi2,frac3pi2,frac3pi4,frac7pi4.

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