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

1 Answer
Nghi N.
Feb 29, 2016

#0, pi/4, pi/2, (3pi)/4, (5pi)/4, (3pi)/2, (7pi)/4 and 2pi#

Explanation:

Call #x/2 = t# . Apply the identity: #sin 2a = 2sin a.cos a#
# (sin t)/(cos t) - 2sin t.cos t = 0#.
#(sin t)/(cos t) - (2sin t.cos t) = 0#
#sin t(1 - 2cos^2 t) = 0#
a. #sin t = sin (x/2) = 0# --> #x/2 = 0, x/2 = pi and x/2 = 2pi# -->
#x = 0 and x = 2pi#
b. #1 - 2cos^2 t = 0# --> #cos^2 t = 1/2# -->
#cos t = cos (x/2) = +-sqrt2/2 #
#cos t = cos (x/2) = sqrt2/2 --> x/2 = +- pi/4 --> x = +-pi/2#
#cos t = cos (x/2) = -sqrt2/2# --> #x/2 = +-(3pi)/4# --> #x = +-3pi/2#
Answers for #(0, 2pi)#
#0, pi/4, pi/2, (3pi)/4, (5pi)/4, (3pi)/2, (7pi)/4, 2pi.#
Note.
#- pi/4# --> #(7pi)/4# (co-terminal)
#-(3pi)/4# --> #(5pi)/4# (co-terminal)

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