# Complex no again

• Dec 11th 2012, 05:10 AM
Tutu
Complex no again
I really need help here!

Using the triple angle identities for sin, cos and tan, Solve the equation
1.) 4x^3 - 3x = -(1/sqrt2)
At first instinct I saw that it resembled the triple angle identity for cosine so I had cos3x=-(1/sqrt2), arccos(1/sqrt2) gave me pi/4 and then cosx is pi/12. However, the answer had cos(11pi/12) and cos(5pi/12), how did they get that?

2.) Solve x^3 - 3sqrt(3)x^2 -3x + sqrt(3) = 0. I couldn't see how it represented any of the triple identities. I thought tan3x is possible but I have problems with the coefficients, they don't tally!

• Dec 11th 2012, 06:34 AM
emakarov
Re: Complex no again
When you are solving an equation like cos(3t) = - 1 / sqrt(2), you should first look at the unit circle and determine the sign and the approximate values of 3t. Use arccos only in the end to get the numerical value, and only if the the argument of cosine is not a multiple of 30 or 45 degrees.

In this case, if you draw a vertical line through x = - 1 / sqrt(2), it intersects the unit circle at 3pi/4 and 5pi/4. So,

$\displaystyle 3t = 3\pi/4+2\pi k$ or
$\displaystyle 3t = 5\pi/4+2\pi k$

where $\displaystyle k\in\mathbb{Z}$. This means

$\displaystyle t = \pi/4 + 2\pi k/3$ or
$\displaystyle t = 5\pi/12 + 2\pi k/3$

Now find all t from 0 to 2pi. Some of them will generate the same solution x = cos(t) to the original equation.
• Dec 11th 2012, 06:41 AM
zhandele
Re: Complex no again
Interesting way to do algebra problems. New to me.

As for the second, I think it does tally up with the triple tangent formula. Looks that way to me. Hope this helps.