1. ## Double angle/sum stuff

sin 3x + cos 2x = 0

I need to solve for x between the interval [0,2pi)

In the previous chapter there was a problem that wanted us to prove that sin 3x = 3cos^2xsinx-sin^3x so i decided to try using that. However i'm not sure if i'm going down the right path. How would you guys do it?

Wolfram Alpha also got

2(pi * n-tan inverse(1+(root 5)+(root 5 +2(root5))) I'm not sure how to get tan out of all that stuff.

~anonymous

2. Hello, RIkT!

Your idea is a good one . . .

We need two Multiple-Angle Identities: . $\begin{array}{ccc}\sin3x &=& 3\sin x - 4\sin^3\!x \\ \cos2x &=& 1 - 2\sin^2\!x \end{array}$

$\text{Solve for }x\!:\;\;\sin 3x + \cos 2x \:=\: 0\;\;\text{ on }[0,\:2\pi]$

We are given: . $\sin x + \cos2x \:=\:0$

Apply identities: . $(3\sin x - 4\sin^3\!x) + (1 - 2\sin^2\!x) \:=\:0$

And we have: . $4\sin^3\!x + 2\sin^2\!x - 3\sin x - 1 \:=\:0$

. . which factors: . $(\sin x + 1)(4\sin^2\!x - 2\sin x - 1) \:=\:0$

We have two equations to solve:

$(a)\;\;\sin x + 1 \;=\:0 \quad\Rightarrow\quad\sin x \:=\:-1 \quad \Rightarrow\quad \boxed{x \:=\:\frac{3\pi}{2}}$

$(b)\;\;4\sin^2\!x - 2\sin x - 1 \:=\:0$
. . Quadratic Formula: . $\sin x \;=\;\frac{2\pm\sqrt{(\text{-}2)^2 - 4(4)(\text{-}1)}}{2(4)} \;=\;\frac{1 \pm\sqrt{5}}{4}$

. . $\text{Then: }\;\sin x \;=\;\frac{1+\sqrt{5}}{4} \quad\Rightarrow\quad x \;=\;\arcsin\left(\frac{1+\sqrt{5}}{4}\right) \quad\Rightarrow\quad\boxed{ x \;=\;\frac{3\pi}{10},\;\frac{7\pi}{10}}$

. . $\text{And: }\;\;\sin x \;=\;\frac{1-\sqrt{5}}{4} \quad\Rightarrow\quad x \;=\;\arcsin\left(\frac{1-\sqrt{5}}{4}\right) \quad\Rightarrow\quad \boxed{x \;=\;\frac{11\pi}{10},\;\frac{19\pi}{10}}$

3. Thanks for the help. I got to the cubic equation. I had trouble factoring it. I got it now. Thank you.