Question #350746

Find x if, 2^sin²x + 2^cos²x = 3


1
Expert's answer
2022-06-15T13:00:48-0400

2sin2x+2cos2x=32^{\sin^2x}+2^{\cos^2x}=3

2sin2x+21sin2x=32^{\sin^2x}+2^{1-\sin^2x}=3

2sin2x+212sin2x=32^{\sin^2x}+2\cdot\frac{1}{2^{\sin^2x}}=3


Let y=2sin2xy=2^{\sin^2x}


y+2y=3y+\frac{2}{y}=3

y23y+2=0y^2-3y+2=0

y1=1y_1=1, y2=2y_2=2


2sin2x=12^{\sin^2x}=1

sin2x=0\sin^2x=0

sinx=0\sin x=0

x=πn,nZx=\pi n, n\in\mathbb{Z}


2sin2x=22^{\sin^2x}=2

sin2x=1\sin^2x=1

sinx=±1\sin x=\pm1

x=(1)nπ2+πn,nZx=(-1)^n\frac{\pi}{2}+\pi n, n\in\mathbb{Z}


Answer:

x=πn,nZx=\pi n, n\in\mathbb{Z}

x=(1)nπ2+πn,nZx=(-1)^n\frac{\pi}{2}+\pi n, n\in\mathbb{Z}





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