How to solve $8sin^2x+4cos^2x=7$ for 0≤x≤2π?

Question

$8sin^2x+4cos^2x=7$ for $0<=x<=2pi$

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Answer 1 · 2017-05-24T23:34:23

Replace $cos^2(x)$ with $1 - sin^2(x)$ :

$8sin^2(x)+4(1-sin^2(x))=7$

Distribute the 4:

$8sin^2(x)+4-4sin^2(x)=7$

Combine like terms:

$4sin^2(x)=3$

Divide both sides by 4:

$sin^2(x) = 3/4$

Use the square root on both sides:

$sin(x) = +-sqrt3/2$

Within the specified domain:

$x = pi/3, 2pi/3, 4pi/3 and 5pi/3$