Hello deej813 Originally Posted by
deej813 Sin2x + _sin2x = 2
1-cosx 1+cosx
i used sin^2x + cos^2 x = 1
then got (sin^2x)/(1-cos^2x)=0
and thats a diff of 2 squares
so its
(sin^2x)/(1-cosx)(1+cosx) = 0
but where does the two come from
what did i do wrong??
I'm not clear what the question is here. My best guess is that you have to solve the equation$\displaystyle \frac{\sin2x}{1-\cos x}+\frac{\sin2x}{1+\cos x}=2$
If that's it then you'd write:$\displaystyle \Rightarrow \frac{\sin2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}=2$
$\displaystyle \Rightarrow \frac{4\sin x \cos x}{1-\cos^2 x}=2$, using $\displaystyle \sin 2x = 2\sin x\cos x$
$\displaystyle \Rightarrow \frac{2\sin x\cos x}{\sin^2x}=1$
$\displaystyle \Rightarrow 2\cot x = 1$
$\displaystyle \Rightarrow \tan x = 2$
$\displaystyle \Rightarrow x = \arctan(2) + n\pi$
Grandad