How do you find all solutions of #cos(x/2)-sinx=0# in the interval #[0,2pi)#?

1 Answer
Nghi N.
Jan 3, 2017

For #(0, 2pi)#
#pi/3, pi, (5pi)/3#

Explanation:

Use trig identity:
#sin x = 2sin (x/2).cos (x/2)#
Solve the equation:
#cos (x/2) - 2sin (x/2).cos (x/2) = 0#
#cos (x/2)((1 - 2sin (x/2) = 0#

a. cos x/2 = 0 --> x/2 = pi/2, and x/2 = (3pi)/2 -->
#x/2 = pi/2 --> x = pi#
#x/2 = (3pi)/2 --> x = 3pi# (rejected because out of area)

b. #(1 - 2sin (x/2) = 0#
Trig table and unit circle -->
#sin (x/2) = 1/2# --> #x/2 = pi/6# and #x/2 = (5pi)/6#
#x/2 = pi/6 --> x = pi/3#
#x/2 = (5pi)/6 --> x = (5pi)/3#
Answers for #(0, 2pi) #pi/3, pi, (5pi)/3#

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