Question 835644: -sin(theta) + cos(2theta) = 0
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! -sin(theta) + cos(2theta) = 0
-sin(theta) + 1 - 2*sin^2(theta) = 0
-z + 1 - 2z^2 = 0 ... let z = sin(theta)
-2z^2 - z + 1 = 0
2z^2 + z - 1 = 0
(z+1)(2z-1) = 0
z+1 = 0 or 2z-1 = 0
z = -1 or z = 1/2
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If z = -1, then
z = -1
sin(theta) = -1
theta = arcsin(-1)
theta = 3pi/2
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If z = 1/2, then
z = 1/2
sin(theta) = 1/2
theta = arcsin(1/2)
theta = pi/6 or theta = 5pi/6
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Answer:
theta = pi/6, theta = 5pi/6, theta = 3pi/2
Note: this is assuming we're restricted to the interval [0, 2pi)
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