Could anyone help me these problems, thanks very much.
Solve the equations:
$\displaystyle i) cos2x = sinx$
$\displaystyle ii) sin2x - 1 = cos2x$
Giving the answers in radians in the range 0 to $\displaystyle 2\pi$
HI
(1) $\displaystyle \cos 2x=\sin x $
$\displaystyle 1-2\sin^2 x =\sin x$
$\displaystyle 2\sin^2 x+\sin x-1=0$
$\displaystyle (2\sin x-1)(\sin x+1)=0$
So $\displaystyle 2\sin x-1=0$ OR $\displaystyle \sin x+1=0 $
Continue from here ..
(2) $\displaystyle \sin 2x-1=\cos 2x$
$\displaystyle 2\sin x\cos x -\cos^2 x -\sin^2 x =\cos^2 x-\sin^2 x$
$\displaystyle 2\cos^2 x-2\sin x\cos x=0$
$\displaystyle \cos^2 x-\sin x\cos x=0$
$\displaystyle \cos x(\cos x-\sin x)=0$
$\displaystyle \cos x=0$ OR $\displaystyle \cos x-\sin x=0\Rightarrow \tan x=1$
Continue from here .