Question 177064: Solve for x on the interval given:
a) 2sin^2x-5cosx+1=0 [0, pie]
b) sinx/2=0 [-pie, pie]
c) cos2x=cos^2x [0, 2pie]
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Solve for x on the interval given:
a) 2sin^2x-5cosx+1=0 [0, pi]
2(1-cos^2) - 5cos + 1 = 0
2-2cos^2 - 5cos + 1 = 0
2cos^2 + 5cos - 3 = 0
Now it's a quadratic in cos(x)
cos(x) = -3, +1/2
Ignore the -3 (it's a complex number)
cos(x) = 1/x
x = 60º, 300º = pi/3 (not pie, pi)
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b) sinx/2=0 [-pi, pi]
Assuming you mean sin(x/2) and not (sin(x))/2:
x/2 = 0, pi, -pi
x = 0
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c) cos2x=cos^2x [0, 2pi]
cos^2 - cos = 0
cos(cos - 1) = 0
cos = 0, cos = 1
For cos(x) = 0:
x = pi/2, 3pi/2
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For cos(x) = 1:
x = 0, 2pi
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