# Thread: solve after sum to product

1. ## solve after sum to product

need values after using sum to product formula.

I used sum to product and get 2sin4xcosx=0
setting each factor to 0, 2sin4x=0 and cosx=0,
I have the answers in the book, but need to know how to get to the answers. x=0, pi/4, pi/2 etc. (adding [(npi)/4] each time.

2. Hello, dashreeve!

Are you really having trouble solving elementary trig equations?

Solve: . $2\sin4x\:=\:0\:\text{ and }\;\cos x\:=\:0$
$2\sin4x \:=\:0 \quad\Rightarrow\quad \sin4x \:=\:0 \quad\Rightarrow\quad 4x \:=\:0 + \pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{4}n \;\;{\color{blue}[1]}$

$\cos x \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{\pi}{2} + \pi n \;\;{\color{blue}[2]}$

But all the solutions of [2] are included in [1].

Therefore, the solution is: . $x \:=\:\frac{\pi}{4}n\;\;\;\text{ for }n \in I$

3. thanks for the encouraging remarks! I'm having a lot of trouble understanding this stuff actually. And half way through the course now, and haven't seen the technique you use there (which is much easier than what I've been doing). Did I miss the rule about sin4x=0 can move to 4x=0+pi(n)??? When was I supposed to learn that? I know my professor has us jumping around the book, but I can't imagine I would forget that little rule..

4. Hello, dashreeve!

Suppose we have: . $2\sin 3x \:=\:1$

Then we have: . $\sin3x \:=\:\frac{1}{2}$

We know that: . $\sin\frac{\pi}{6} \:=\:\frac{1}{2}\:\text{ and }\:\sin\frac{5\pi}{6} \:=\:\frac{1}{2}$

So the angle is either $\frac{\pi}{6}\,\text{ or }\,\frac{5\pi}{6}$ . . . plus some multiple of $2\pi.$

So we have: . $3x \;=\;\begin{Bmatrix}\dfrac{\pi}{6} + 2\pi n \\ \\[-3mm]\dfrac{5\pi}{6} + 2\pi n \end{Bmatrix}$

Therefore: . $x \;=\;\begin{Bmatrix}\dfrac{\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{5\pi}{18} + \dfrac{2\pi}{3}n \end{Bmatrix}$

5. I see. Thanks for your help on several questions. It's really teaching me a lot.