Diffcult Trigonometry Equation Solving

**shinn** · September 13th 2008, 12:48 AM

Could anyone help me with this problem that I've been struggling with?

Find all x such that:

$cos x + sin x = 1 + sin 2x$ , for 0 \< x \< 2x where \< is greater than or equal to.

Thanks in advance.

**thelostchild** · September 13th 2008, 01:07 AM

square both sides of the equation
using some identities it simplifies down to

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

solve that then test all the solutions you got in the original equation (squaring both sides causes some extra solutions to pop up)

**Moo** · September 13th 2008, 01:09 AM

Hello,

Originally Posted by shinn

Could anyone help me with this problem that I've been struggling with?

Find all x such that:

$cos x + sin x = 1 + sin 2x$ , for 0 \< x \< 2x where \< is greater than or equal to.

Thanks in advance.

Please note that $(\sin(x)+\cos(x))^2=\sin^2(x)+\cos^2(x)+2 \sin(x) \cos(x)=1+\sin(2x)$

So your equation is :

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

$1+\sin(2x)=(1+\sin(2x))^2$

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

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

**shinn** · September 13th 2008, 01:26 AM

Wow that is neat...... never thought of squaring when trying to eliminate the 1............

Thanks greatly for your help.

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