How do you solve #cos 2 x = sin x# over the interval 0 to 2pi?

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How do you solve #cos 2 x = sin x# over the interval 0 to 2pi?

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Answer 1 · 2016-04-16T02:48:09

#pi/6, (5pi)/6 and (3pi)/2#

Explanation:

Apply the trig identity: #cos 2x = 1 - 2sin^2 x.# We get:
#1 - 2sin^2 x = sin x#
# - 2cos^2 x - sin x + 1 = 0.#
Solve this quadratic equation for sin x.
Since a - b + c = 0, use shortcut. The 2 real roots are; -1 and
#-c/a = 1/2.#
Trig unit circle and trig table give -->
a. sin x = -1 --># x = (3pi)/2.#
b. #sin x = 1/2# --> #x = pi/6# and #x = pi - pi/6 = (5pi)/6#

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How do you solve #cos 2 x = sin x# over the interval 0 to 2pi?

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