cos(x)+sin(-3x)=0 - Doubt in a particular step

[Mathallica]

Hello!

It maybe a stupid question, I have these two equations in which I don't exacly understand a particular step
The domain is from (-pi/2 to pi)


cos(x)+sin(-3x)= 0

<=> cos(x)=-sin(-3x)

<=> cos(x)=sin(3x)

<=> sin(pi\2-x) = sin(3x) ( because cos y = sin (pi/2 -y) )

<=> pi/2-x= 3x+2*k*pi V pi/2-x = pi - 3x + 2*k*pi , k c Z. - It is this particular step i don't really get. :\

Can someone explain it to me?



There's another as well:

cos(2x)=-sin(2x)

<=>2x=-pi\4 + k*pi., k c Z.


Which trigonometric property allows this?

Thank you in advance for you time and help!

 

[ksdhart2]

So, I'm assuming that the V is meant to be the OR operator. If that's not right, please reply with the necessary correction. Assuming it is, the preceding step is:

\(\displaystyle sin(\dfrac{\pi}{2}-x)=sin(3x)\)

Taking the inverse sine of both sides gives us:

\(\displaystyle \dfrac{\pi}{2}-x=3x\)

Then, because \(\displaystyle sin(x)=sin(x)+2k\pi \text{ for } k \in \mathbb{Z}\), we have to account for all the solutions. That's the first half of the step you're confused about. The second half comes about because of another identity:

\(\displaystyle sin(\pi-x)=sin(x)\)

Applying that to the starting equation gives us:

\(\displaystyle sin(\dfrac{\pi}{2}-x)=sin(\pi-3x)\)

Again, take the inverse sine of both sides and account for all the solutions, and there's the second half.

 

[Mathallica]

Many thanks, ksdhart2! I see it all now! Sorry for the late reply!

Cheers and thank you again for you time