1.

Solve for θ :2sin'e+ sin2θ-23-2cosO _ 4sin0-cos2θ + sin28-0.dyOR

Answer»

2sin^2x+ (2sinx.cosx)^2=2

2sin^2x+4sin^2x.cos^2x=2

sin^2x+2sin^2x.cos^2x=1

2sin^2x.cos^2x= 1-sin^2x

2sin^2x.cos^2x= cos^2x

2sin^2x=1

sin^2x= 1/2

sinx= 1/√2

=> x= pi/4 or 45 degrees


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