How do you solve #2sin^2a=2+cosa# and find all solutions in the interval #[0,2pi)#?

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How do you solve #2sin^2a=2+cosa# and find all solutions in the interval #[0,2pi)#?

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Answer 1 · 2016-09-15T03:39:59

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

Explanation:

Replace in the equation #2sin^2 a# by #2(1 - cos^2 a)#, we get:
#2 - 2cos^2 a = 2 + cos a#
#-2cos^2 a - cos a = 0#
#- cos a(2cos a + 1) = 0#
Trig table and unit circle -->
a. #cos a = 0# --> #a = pi/2# and #a = (3pi)/2#
b. 2cos a + 1 = 0
#cos a = -1/2# --> #a = +- (2pi)/3#
The arc# (-2pi)/3# and arc #(4pi)/3# are co-terminal
Answers for #(0, 2pi):#
#pi/2; (2pi)/3, (4pi)/3, (3pi)/2#

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How do you solve #2sin^2a=2+cosa# and find all solutions in the interval #[0,2pi)#?

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