# solve the equations

• Oct 27th 2009, 07:14 AM
srobrien
solve the equations
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 $\displaystyle 2\pi$

• Oct 27th 2009, 07:21 AM
Quote:

Originally Posted by srobrien
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 .
• Oct 27th 2009, 07:45 AM
srobrien
Thank you
• Oct 27th 2009, 09:14 AM
james_bond
Quote:

HI

(1) $\displaystyle \cos 2x=\sin x$

$\displaystyle 2\sin x\cos x =\sin x$

Are you sure?
• Oct 28th 2009, 12:28 AM