Hay,

I started with co-functions a little while back, I got the concept and understand the A + B = 90^{0},

But i got this one that threw me a curve-ball,

sin (90^{0}+x) = 2 cos^{2}x

I spoke to my lecturer about this one and he said just to treat the second part as a normal square,

Where is what i tried...

sin (90^{0}+x) = 2[sin(90^{0}-x)^{2}]

sin (90^{0}+x) = 2[(90^{0}-x)(90^{0}-x)]

90^{0}-x= 2[810^{0}- 180^{0}x+x^{2}]

90^{0}-x= 1620^{0}- 360^{0}x+ 2x

2x^{2}- 361^{0}x+ 1530^{0}= 0

Using quadratic formula:

.'.x= 3 + "root 12231" /4 or .'.x= 3 -"root 12231" /4

.'.x= 28.398 or .'.x= -27.145

Did i miss something here or look at it from a wrong angle?