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.

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- May 2nd 2006, 04:11 PMCONFUSED_ONETrig. EquationOKay, 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 PMJameson
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 $\displaystyle \frac{\pi}{2}$ and it will do so at every at every $\displaystyle \pi$ after that. So I would write one zero as $\displaystyle \frac{\pi}{2}+k\pi$ where k is any integer.

- May 2nd 2006, 04:30 PMCONFUSED_ONE
Oh yes, silly me. I forgot about cosx=0. Thank you very much Jameson!