# Thread: Find all primary trig solutions

1. ## Find all primary trig solutions

"Find all primary solutions in
sin6(theta)+sin3(theta) = 0

2. or at least tell me how to simplify or alter that equation, thanks

3. Originally Posted by erockdontstop

"Find all primary solutions in
sin6(theta)+sin3(theta) = 0

$\displaystyle sin(6x) + sin(3x) = 0$

$\displaystyle sin(6x) = - sin(3x)$

$\displaystyle \frac{sin(6x)}{-sin(3x)} = 1$

$\displaystyle -2 cos(3x) = 1$

$\displaystyle cos(3x) = -\frac{1}{2}$

$\displaystyle 3x = arccos(-\frac{1}{2})$

$\displaystyle 3x = \frac{2\pi}{3}$

$\displaystyle x = \frac{2 \pi}{9}$

Now plug x in to the original equation to see if it equals 0

$\displaystyle sin(6(\frac{2\pi}{9}) + sin(3*(\frac{2\pi}{9}) = ?$

$\displaystyle -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0$

remember that there are more solutions to this problem such as
$\displaystyle x = \pi,\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{2 \pi}{9},\frac{2 \pi}{9}, \frac{4 \pi}{9}$ and so on

4. [quote=fonso_gfx;238504]$\displaystyle sin(6x) + sin(3x) = 0$

$\displaystyle \frac{sin(6x)}{-sin(3x)} = 1$

$\displaystyle -2 cos(3x) = 1$

how did you that?

5. [quote=erockdontstop;238509]
Originally Posted by fonso_gfx
$\displaystyle sin(6x) + sin(3x) = 0$

$\displaystyle \frac{sin(6x)}{-sin(3x)} = 1$

$\displaystyle -2 cos(3x) = 1$

how did you that?
I used the double angle formula

$\displaystyle sin(6x) = sin(2(3x)) = 2sin(3x)*cos(3x)$

Now when when you divide by -sin(3x) you get only

$\displaystyle -2cos(3x)$