Trigonometry equation 2

**dorkymichelle** · December 8th 2009, 09:36 PM

$4cos^23x=1$
$cos^23x=1/4$
$(cos^22\pi)+cos^2x=1/4$ can i do a square root here?
assuming I cant I went on and did
$(2cos^2x-1)^2+cos^2x=1/4$
[tex](4cos^4x-2cos^2x+1)+cos^2x=1/4[\math]
$4cos^4x-cos^2x+1=1/4$
$cos^2x(4cos^2x-1)+1=1/4$

how .. do I do this.. especially without the unit circle
I looked at the answers and the values are not on the unit circle.

**11rdc11** · December 8th 2009, 10:00 PM

Originally Posted by dorkymichelle

$4cos^23x=1$
$cos^23x=1/4$
$(cos^22\pi)+cos^2x=1/4$ can i do a square root here?
assuming I cant I went on and did
$(2cos^2x-1)^2+cos^2x=1/4$
[tex](4cos^4x-2cos^2x+1)+cos^2x=1/4[\math]
$4cos^4x-cos^2x+1=1/4$
$cos^2x(4cos^2x-1)+1=1/4$

how .. do I do this.. especially without the unit circle
I looked at the answers and the values are not on the unit circle.

Here is how I would approach the problem

$4\cos^2{3x}=1$

$\cos^2{3x}=\frac{1}{4}$

now take the square root

$\cos{3x} = -\frac{1}{2}$ or $\cos{3x} = \frac{1}{2}$

Can you finish up from here?

**dorkymichelle** · December 8th 2009, 10:19 PM

The next step I got is
$cos2x+cosx=1/2$
[tex]2cos^2x-1+cosx=1/2
$2cos^2x+cosx-3/2=0$
$cosx(2cosx+1)-2/3=0$
$cosx = 1/2$
$so \pi/3 + n\pi$
but the book's answer is $\pi/9+2\pi n/3, 5\pi/9+2\pi n/3,2\pi/9+2\pi n/3, 4\pi/9+2\pi n/3$

How do I get all these answers?!

**Prove It** · December 8th 2009, 11:06 PM

$\cos{(3x)}$ does NOT equal $\cos{x} + \cos{(2x)}$ .

So far you have two possibilities:

Case 1:

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

Cosine is positive in the first and fourth quadrants, and to get to $x$ , you have to "undo" the cosine - so take the arccos (or $\cos^{-1}$ ...

$3x = \arccos{\frac{1}{2}}$

$3x = \left \{ \frac{\pi}{3}, 2\pi - \frac{\pi}{3} \right \} + 2\pi n, n \in \mathbf{Z}$ , since the period of cosine is $2\pi$ .

$3x = \left \{ \frac{\pi}{3}, \frac{5\pi}{3} \right \} + 2\pi n$

$x = \left \{ \frac{\pi}{9}, \frac{5\pi}{9} \right \} + \frac{2\pi n}{3}$ .

Case 2:

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

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

$3x = \left \{ \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3} \right \} + 2\pi n$ , since cosine is negative in the second and third quadrants and the period is $2\pi$

$3x = \left \{ \frac{2\pi}{3}, \frac{4\pi}{3} \right \} + 2\pi n$

$x = \left \{ \frac{2\pi}{9}, \frac{4\pi}{9} \right \} + \frac{2\pi n}{3}$ .

So putting it all together, the possibilities are

$\left \{ \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9} \right \} + \frac{2\pi n}{3}$ .

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