Re: HSC 2013 4U Marathon
Sub x=cosa, \therefore cos^{2}a+cos^{2}3a=1\Rightarrow cos2a+cos6a=0\therefore 2cos2acos4a=0
\therefore 2a=\frac{\pi }{2}, \frac{3\pi }{2} or 4a=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2} $we get$ x=\pm \frac{\sqrt{2}}{2}and\pm \sqrt{\frac{2\pm...