## anonymous one year ago what steps would I take to solve the problem: Solve each equation on the interval 0<alpha<2pie 1)2sinx+sqrt2=0 2)2sin^2x-3sinx+1=0

1. anonymous

Do you mean $$0<x<2\pi$$ or do you mean $$0 < \alpha < 2\pi$$?

2. anonymous

the 2nd one

3. anonymous

But there are no alphas in your problems...

4. anonymous

that why im confused as well lol maybe it is the first one

5. anonymous

Let's just assume that it is the first one, because that would make more sense lol

6. anonymous

So, for the first one.$2\sin(x)+\sqrt{2}=0$$2\sin(x)=-\sqrt{2}$$\sin(x)=-\frac{\sqrt{2}}{2}$Now, use your trig tables to find what values of x satisfy that while making sure $$0 < x < 2\pi$$. Get it?

7. anonymous

oh so you pretty much want to get the sin alone

8. anonymous

what if there is sin and cos in the problem. ex:8-12sin^2x=4cos^2x

9. anonymous

Yeah. When solving trig equations, you want to (1) get all the trig stuff into 1 trigonometric function. So, if you have both a sine and a cosine or something, you want to (usually) get it all in terms of sine or cosine. (2) You want to solve for the trig function.

10. anonymous

For that kind of thing, that's where you have to use your identities. For instance, in that question you just posted, trying using your pythagorean identities. $\implies8 - 12\sin^2(x)=4\cos^2(x)$$\implies8-12\sin^2(x)=4(1-\sin^2(x))$$\implies0=8\sin^2(x)-4$$\implies\frac12=\sin^2(x)$Does this make sense?

11. anonymous

ahh yes I finally understand :D thanks so much!

12. anonymous

You're welcome. :)