Question 835644
-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)