Find x if, 2^sin²x + 2^cos²x = 3
2sin2x+2cos2x=32^{\sin^2x}+2^{\cos^2x}=32sin2x+2cos2x=3
2sin2x+21−sin2x=32^{\sin^2x}+2^{1-\sin^2x}=32sin2x+21−sin2x=3
2sin2x+2⋅12sin2x=32^{\sin^2x}+2\cdot\frac{1}{2^{\sin^2x}}=32sin2x+2⋅2sin2x1=3
Let y=2sin2xy=2^{\sin^2x}y=2sin2x
y+2y=3y+\frac{2}{y}=3y+y2=3
y2−3y+2=0y^2-3y+2=0y2−3y+2=0
y1=1y_1=1y1=1, y2=2y_2=2y2=2
2sin2x=12^{\sin^2x}=12sin2x=1
sin2x=0\sin^2x=0sin2x=0
sinx=0\sin x=0sinx=0
x=πn,n∈Zx=\pi n, n\in\mathbb{Z}x=πn,n∈Z
2sin2x=22^{\sin^2x}=22sin2x=2
sin2x=1\sin^2x=1sin2x=1
sinx=±1\sin x=\pm1sinx=±1
x=(−1)nπ2+πn,n∈Zx=(-1)^n\frac{\pi}{2}+\pi n, n\in\mathbb{Z}x=(−1)n2π+πn,n∈Z
Answer:
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