sin (x) = cos (2x) x = ??

 

$$\begin{array}{rlll} sin(x)&=&cos(2x)\\ sin(x)&=&cos^2(x)-sin^2(x)\\ sin(x)&=&1-sin^2(x)-sin^2(x)\\ sin(x)&=&1-2sin^2(x)\\ 2sin^2(x)+sin(x)-1&=&0\qquad &\mbox{let y=sin(x)}\\ 2y^2\:\:+y\:\:-1&=&0\\ 2y^2\:\:+2y-1y\:\:-1&=&0\\ 2y(y+1)-1(y+1)&=&0\\ (2y-1)(y+1)&=&0\\ \end{array}$$

$$\begin{array}{rllrl} 
 2y-1&=&0\qquad y+1&=&0\\ y&=&\frac{1}{2}\qquad y&=&-1\\ sin(x)&=&\frac{1}{2}\qquad sin(x)&=&-1 \end{array}$$


$$x=\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{3\pi}{2}\qquad \mbox{for}\:\: 0\le x\le 2\pi}}$$

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sin(x)=cos(2x)

sin(x)-cos(2x)=0                 | Move x to one side of equation

sin(x)-(1-sin²(x))=0             | Double angle identity

sin²(x)+sin(x)-1=0              | simplify terms

(2sin(x)-1)(sin(x)+1)=0       |

x=π/6 , 3π/2                       |   if you need degrees, x=30º ,270º

Anonymous left a 2 out of the third and fourth lines above.  Should be

sin(x)-(1-2sin²(x))=0             | Double angle identity

2sin²(x)+sin(x)-1=0              | simplify terms

The 2 is there correctly in the fifth line though!

Could also have x=5π/6 (150°) as a second quadrant solution.