1. ## Quadratic trig equation #2

and

$\displaystyle 2sin^3(3x) - sin^2(3x) - 2sin(3x) +1 = 0$

I got down half of it $\displaystyle 3x = arcsin(1/2)$

The other half is what is confusing me. How do I get the values for x on this one? $\displaystyle 3x = arcsin(+/- sqrt(1))$

I know arcsin is opposite / hypotenuse.

2. Originally Posted by FaRmBoX
and

$\displaystyle 2sin^3(3x) - sin^2(3x) - 2sin(3x) +1 = 0$

I got down half of it $\displaystyle 3x = arcsin(1/2)$

The other half is what is confusing me. How do I get the values for x on this one? $\displaystyle 3x = arcsin(+/- sqrt(1))$

I know arcsin is opposite / hypotenuse.
$\displaystyle 2\sin^3(3x) - \sin^2(3x) - 2\sin(3x) +1 = 0$

This time solve the cubic

$\displaystyle 2a^3 - a^2 - 2a +1 = 0$

what values for $\displaystyle a$ did you get?

How did you go with the first question you posted?

3. I got $\displaystyle a = +/- squareroot(1)$ and $\displaystyle a = 1/2$

4. $\displaystyle 3x = \arcsin(1/2)$

$\displaystyle 3x = \frac{\pi}{6}$

$\displaystyle x = \frac{\pi}{18}$

Also

$\displaystyle \sin(3x) = \pm \sqrt{1}$

$\displaystyle \sin(3x) = -1,1$

Can you take it from here?

5. For $\displaystyle 3x = \arcsin(1/2)$

I also got $\displaystyle 3x = \frac{5\pi}{6}$
$\displaystyle x = \frac{5\pi}{18}$

Is that right?

For the bottom, did I understand this correctly? $\displaystyle \pm \sqrt{1} = -1,1$

6. Originally Posted by FaRmBoX
For $\displaystyle 3x = \arcsin(1/2)$

I also got $\displaystyle 3x = \frac{5\pi}{6}$
$\displaystyle x = \frac{5\pi}{18}$

Is that right?
Yep, as sin has more than one solution for that value.

The best solution will look a little like

$\displaystyle x = \frac{(4n+1)\pi}{18}, n=0,1,2\dots$

Originally Posted by FaRmBoX
For [tex]
For the bottom, did I understand this correctly? $\displaystyle \pm \sqrt{1} = -1,1$
Yep the square root of 1 is -1 and 1.