Quadratic trig equation #2

**FaRmBoX** · January 4th 2010, 04:17 PM

and

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

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

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

I know arcsin is opposite / hypotenuse.

**pickslides** · January 4th 2010, 04:23 PM

Originally Posted by FaRmBoX

and

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

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

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

I know arcsin is opposite / hypotenuse.

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

This time solve the cubic

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

what values for $a$ did you get?

How did you go with the first question you posted?

**FaRmBoX** · January 4th 2010, 04:30 PM

I got $a = +/- squareroot(1)$ and $a = 1/2$

**pickslides** · January 4th 2010, 04:36 PM

$3x = \arcsin(1/2)$

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

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

Also

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

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

Can you take it from here?

**FaRmBoX** · January 4th 2010, 04:46 PM

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

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

Is that right?

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

**pickslides** · January 4th 2010, 05:38 PM

Originally Posted by FaRmBoX

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

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

Is that right?

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

The best solution will look a little like

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

Originally Posted by FaRmBoX

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

Yep the square root of 1 is -1 and 1.

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