# Math Help - Diffcult Trigonometry Equation Solving

1. ## Diffcult Trigonometry Equation Solving

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.

2. 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)

3. 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.

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 :

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

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

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

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

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

Thanks greatly for your help.