How do you solve #2sin^2 x=2+cos x# between 0 and 2pi?

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How do you solve #2sin^2 x=2+cos x# between 0 and 2pi?

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Answer 1 · 2016-03-09T06:19:05

#pi/2, (2pi)/3, (4pi)/3, and (3pi)/2#

Explanation:

Replace in the equation #sin^2 x# by #(1 - cos^2 x#
#2(1 - cos^2 x) = 2 + cos x#
#- 2cos^2 x = cos x#
#cos x + 2cos^2 x = cos x(1 + 2cos x) = 0#
a. cos x = 0 --> #x = pi/2 and x = (3pi)/2#
b. (1 + 2cos x) = 0 --> #cos x = - 1/2# --># x = +- (2pi)/3#
The co-terminal arc of - (2pi)/3 is arc (4pi)/3.
Answer for #(0, 2pi)#:
#pi/2, (2pi)/3, (4pi)/3, (3pi)/2 #

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How do you solve #2sin^2 x=2+cos x# between 0 and 2pi?

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