How do you solve # cos (x/2) - sinx = 0# over the interval 0 to 2pi?

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How do you solve # cos (x/2) - sinx = 0# over the interval 0 to 2pi?

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Answer 1 · 2016-02-17T23:50:35

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

Explanation:

Apply the trig identity: sin 2a = 2sin a.cos a
Replace sin x by #2sin (x/2).cos (x/2)# -->
#cos (x/2) - 2sin (x/2).cos (x/2) = cos (x/2)(1 - 2sin (x/2) )= 0#
On the trig unit circle,
a. cos x = 0 --> arc #x = +- pi/2#.
#x = pi/2 and x = (3pi)/2# (co-terminal with# -pi/2#)
b. #sin x = 1/2# --> arc #x = pi/6# and #x = (5pi)/6.#

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How do you solve # cos (x/2) - sinx = 0# over the interval 0 to 2pi?

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