How do you solve #[(1+cosx)/(sinx)]=-1# over [0,2pi)?

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Answer 1 · 2015-04-17T03:33:48

Here is one way to solve it, by using these trig identities

#f(x) = (1 + cos x) + sin x = 0#. Change variable #x# to #x/2#.

#f(x) = 2*cos^2 (x/2) + 2*sin (x/2)*cos (x/2) = 0#

#f(x) = 2cos (x/2)*(cos (x/2) + sin (x/2) = 0#

Next, solve:

#cos (x/2) = 0 --> x/2 = pi/2 and x/2 = 3pi/2 --> x = pi and x = (3pi)/2.#

#(sin x + cos x) = sqrt2*sin (x/2 + pi/4) = 0#

#sin (x/2 + pi/4) = 0 --> x/2 + pi/4 = pi and 2pi #

#x/2 = pi - pi/4 = (3pi)/4 --> x = (6pi)/4 = (3pi)/2#

#x/2 = 2pi - pi/4 --> x/2 = (7pi)/4 -->x = (14pi)/4 = (7pi)/2# (out of interval)
Answers within (#0, 2pi#): #pi#; and #(3pi)/2#

Check:
When #x = pi --> cos x = -1; sin x = 0 --> f(x) = 1 - 1 = 0# Correct
When #x = (3pi)/2 --> cos x = 0; sin x = -1--> f(x) = 1 - 1 = 0# Correct