Asked by Seven
Solve 2 + cos^2(theta)=3sin^2(theta) for 0=< theta < 360
Answers
Answered by
Thales
We know that cos^2(theta) + sin^2(theta) = 1.
So, using that,
2 + cos^2(theta)=3sin^2(theta)
2 + cos^2(theta)=3(1 - cos^2(theta))
2 + cos^2(theta)=3 - 3cos^2(theta))
4cos^2(theta)= 1
cos^2 (theta)=1/4
cos (theta)=+-(1/4)^1/2
theta = arccos(1/2) or theta = arccos(-1/2)
theta = 60° or 300° or theta = 120° or 240°
So, using that,
2 + cos^2(theta)=3sin^2(theta)
2 + cos^2(theta)=3(1 - cos^2(theta))
2 + cos^2(theta)=3 - 3cos^2(theta))
4cos^2(theta)= 1
cos^2 (theta)=1/4
cos (theta)=+-(1/4)^1/2
theta = arccos(1/2) or theta = arccos(-1/2)
theta = 60° or 300° or theta = 120° or 240°
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