Solving a trigonometric equation

[angel_4_ever3]

I'm trying to solve:

sin2x = -cos2x

so my first step was...

2sinxcosx = -cos^2x - sin^2x

and then...

2sinxcosx + cos^2x + sin2x = 0

but I'm confused on how I would end up factoring 2sinxcosx + cos^2x + sin2x, or if I did something wrong earlier in the problem...thank you

 

[symbolipoint]

Can you find another trigonometric identity which would allow you to express the original equation in terms of either sine, or cosine, but not both? Maybe a different double angle identity?

 

[HallsofIvy]

I'm trying to solve:

sin2x = -cos2x

so my first step was...

2sinxcosx = -cos^2x - sin^2x

and then...

2sinxcosx + cos^2x + sin2x = 0

but I'm confused on how I would end up factoring 2sinxcosx + cos^2x + sin2x, or if I did something wrong earlier in the problem...thank you

Why not just solve siny= -cos y first and then find x?

 

[CaptainADHD]

Why not just solve siny= -cos y first and then find x?

I'm bad with identities, but wouldn't that be 3*pi/4?

square root of two over two equals negative negative of the same.