SOLUTION: Solve, giving all solutions between 0 and 360 degrees: {{{cot(z/2)}}} - {{{2cos(z/2)}}} = 0

Algebra ->  Trigonometry-basics -> SOLUTION: Solve, giving all solutions between 0 and 360 degrees: {{{cot(z/2)}}} - {{{2cos(z/2)}}} = 0      Log On


   

Question 453840: Solve, giving all solutions between 0 and 360 degrees:
cot%28z%2F2%29 - 2cos%28z%2F2%29 = 0
Answer by poliphob3.14(115) About Me  (Show Source):
You can put this solution on YOUR website!
First we simplify the equation:cos%28z%2F2%29%2Fsin%28z%2F2%29-2cos%28z%2F2%29=0, multiply both
sides by sin%28z%2F2%29 and get:cos%28z%2F2%29-2sin%28z%2F2%29%2Acos%28z%2F2%29=0, factor
further, cos%28z%2F2%29%281-2sin%28z%2F2%29%29=0. This product will be zero if:
cos%28z%2F2%29=0 or 1-2sin%28z%2F2%29=0. Solving these two equivalent equations
we find:1)cos%28z%2F2%29=0=>z%2F2=%28pi%29%2F2=>z=pi and z%2F2=3%2Api%2F2=>
z=3%2Api.
2)1-2si%28z%2F2%29=0=>sin%28z%2F2%29=1%2F2=>z%2F2=pi%2F6=>z=pi%2F3 and,
z%2F2=5%2Api%2F6=>z=5%2Api%2F3