# Trig. Equation

• May 2nd 2006, 04:11 PM
CONFUSED_ONE
Trig. Equation
OKay, so how would this equation be solved,
cosx-2sinxcosx=0

so I've tried this way...
cosx-2sinxcosx=0
cosx(1-2sinx)=0
and after that, i'm stuck.
• May 2nd 2006, 04:19 PM
Jameson
Well this has multiple answers now doesn't it? :) You've factored correctly. You know that you're equation will be zero when cos(x)=0 and when 1-2sin(x)=0. So solve for both. I'll start you. The first time cos(x)=0 starting from 0 and moving positively is at $\frac{\pi}{2}$ and it will do so at every at every $\pi$ after that. So I would write one zero as $\frac{\pi}{2}+k\pi$ where k is any integer.
• May 2nd 2006, 04:30 PM
CONFUSED_ONE
Oh yes, silly me. I forgot about cosx=0. Thank you very much Jameson!