MHB How would you solve for x using the double angle formula?

[bsmithysmith]

$$2sin(2x)-3sin(x)=0$$

We did this in my class, but there were some parts where I was really confused. I know that we need to use the double angle formula, and the double angle formula for Sine is:

$$sin(2x) = 2sin(x)cos(x)$$

and correct me if I'm wrong. So what I had down is:

$$2(2sin(x)cos(x))-3sin(x)=0$$

From thereon, I don't know where I should distribute it or something else. I checked Wolframalpha (just in case), and it showed:

$$4cos(x)sin(x)-3sin(x)=0$$

I would think that the 2 would just distribute evenly, but how would I properly get through this part?

 

[MarkFL]

You are proceeding correctly. We are given:

$$2\sin(2x)-3\sin(x)=0$$

Applying the double-angle identity for sine, we obtain:

$$2\left(2\sin(x)\cos(x)\right)-3\sin(x)=0$$

$$4\sin(x)\cos(x)-3\sin(x)=0$$

Now what you want to do is factor, and use the zero-factor property...

 

[bsmithysmith]

You are proceeding correctly. We are given:

$$2\sin(2x)-3\sin(x)=0$$

Applying the double-angle identity for sine, we obtain:

$$2\left(2\sin(x)\cos(x)\right)-3\sin(x)=0$$

$$4\sin(x)\cos(x)-3\sin(x)=0$$

Now what you want to do is factor, and use the zero-factor property...

So I was correct after all (outside the forum)!

$$4\sin(x)\cos(x)-3\sin(x)=0$$
$$sin(x)(4cos(x)-3)=0$$

so it's $$sin(x)=0$$
and $$cos(x)=3/4$$

For sine, $$x=0, pi$$ and for cosine, it's $$2pi-cos^1(3/4)$$ and $$cos^-1(3/4)$$ Note, it's Cosine inverse

 

[MarkFL]

Yes, those are the solutions if you are restricted to:

$$0\le x<2\pi$$

To denote special characters using $\LaTeX$ like $$\pi$$ and functions with fractions like $$\cos^{-1}\left(\frac{3}{4}\right)$$, precede them with backslashes and use the frac command like so:

[noparsetex]$$\pi$$[/noparsetex]

[noparsetex]$$\cos^{-1}\left(\frac{3}{4}\right)$$[/noparsetex]