solve the equations

**srobrien** · October 27th 2009, 07:14 AM

Could anyone help me these problems, thanks very much.

Solve the equations:

$i) cos2x = sinx$

$ii) sin2x - 1 = cos2x$

Giving the answers in radians in the range 0 to $2\pi$

**mathaddict** · October 27th 2009, 07:21 AM

Originally Posted by srobrien

Could anyone help me these problems, thanks very much.

Solve the equations:

$i) cos2x = sinx$

$ii) sin2x - 1 = cos2x$

Giving the answers in radians in the range 0 to $2\pi$

HI

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

$1-2\sin^2 x =\sin x$

$2\sin^2 x+\sin x-1=0$

$(2\sin x-1)(\sin x+1)=0$

So $2\sin x-1=0$ OR $\sin x+1=0$

Continue from here ..

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

$2\sin x\cos x -\cos^2 x -\sin^2 x =\cos^2 x-\sin^2 x$

$2\cos^2 x-2\sin x\cos x=0$

$\cos^2 x-\sin x\cos x=0$

$\cos x(\cos x-\sin x)=0$

$\cos x=0$ OR $\cos x-\sin x=0\Rightarrow \tan x=1$

Continue from here .

**srobrien** · October 27th 2009, 07:45 AM

Thank you

**james_bond** · October 27th 2009, 09:14 AM

Originally Posted by mathaddict

HI

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

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

Are you sure?

**mathaddict** · October 28th 2009, 12:28 AM

Originally Posted by james_bond

Are you sure?

Edited . THanks for pointing out , ok now ?

**james_bond** · October 28th 2009, 04:41 AM

Yeah (I hope). Thanks for correcting.

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