1. ## Trigonometry equation 2

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

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

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

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

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

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

now take the square root

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

Can you finish up from here?

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

How do I get all these answers?!

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

So far you have two possibilities:

Case 1:

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

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

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

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

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

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

Case 2:

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

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

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

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

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

So putting it all together, the possibilities are

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