Question

$2^{cos2x} =3* 2^{ cos^{2}x } -4$

Answer 1

[tex] 2^{cos2x} = 3*2^{cos2x}- 2^{2} - 2*2^{cos2x} =- 2^{2} 2^{cos2x}= \frac{- 2^{2} }{-2} 2^{cos2x}=2 2^{cos2x}= 2^{1} cos2x=1 1 - 2 sin^{2} (x)=1 2 sin^{2} (x)=0 sin^{2} (x)=0 sinx=0 x= \pi =180 grade [/tex]