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#