PLEASE HELP ASAP (PRECALC)

1. continued

$cos(2\theta)=(0.5+sin\theta)^2-sin^2\theta\\ cos(2\theta)=(0.25+sin\theta+sin^2\theta)-sin^2\theta\\ cos(2\theta)=0.25+sin\theta\\ or\\ cos(2\theta)=0.25+cos\theta-0.5\\ cos(2\theta)=cos\theta-0.25\\ $

$cos2\theta=cos^2\theta-sin^2\theta\\ cos2\theta=1-2sin^2\theta\\ so\\ 1-2sin^2\theta=0.25+sin\theta\\ 2sin^2\theta+sin\theta-0.75=0\\ let\;\; x=sin\theta\\ 2x^2+x-0.75=0\\ x=\frac{-1\pm\sqrt{1+6}}{4}\\ x=\frac{-1\pm\sqrt{7}}{4}\\$

x must be positive so

$sin\theta = \frac{\sqrt7 -1 }{4}\\ sin^2\theta=\frac{7+1-2\sqrt7}{16}\\ sin^2\theta=\frac{8-2\sqrt7}{16}\\ cos^2\theta=1-\frac{8-2\sqrt7}{16}\\ sin^2\theta-cos^2\theta = \frac{8-2\sqrt7}{16}-(1-\frac{8-2\sqrt7}{16})\\ sin^2\theta-cos^2\theta = 2*\frac{8-2\sqrt7}{16}-1\\ sin^2\theta-cos^2\theta = \frac{4-\sqrt7}{4}-1\\ sin^2\theta-cos^2\theta = \frac{-\sqrt7}{4}\\ cos(2\theta)= \frac{-\sqrt7}{4}\\$

I have not checked this AT ALL. So you better check carefully for stupid mistakes.

Oooooohhhh Melody, you were SOOOOOO CLOOOOSEEEE!

So actually, the solution that I found out is that you actually have to move the sin to the other side and leave the 1/2 on the other. See if you can solve from there. Your solution takes so much more work btw lol!

Guest, Why would I want to solve it from anywhere?

If there are any errors in my solution they are only careless ones.

I was only answering for your benefit not for mine. 

Do you realize how rude your comment sounds?

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Thanks for the point Chris.