Solving trigonometric equations?

Apr 2012
72
0
Seattle, Washington
Hi, I'm working on this problem, and it's throwing me off a bit because it's a little different from what I've been doing.

Solve 4(cos^2(3x)) 3 = 0 for [0, 2π). Give exact answers.

Here is what I've done so far:

4 (cos^2(3x)) - 3 = 0

4 (cos^2(3x)) = 3 (added 3)

cos^2(3x) = 3/4 (divided by 4)


The three is throwing me off a little and I don't know what the next step is.

Please let me know if I can give anymore information or make anything clear.
 

topsquark

Forum Staff
Jan 2006
11,575
3,454
Wellsville, NY
Hi, I'm working on this problem, and it's throwing me off a bit because it's a little different from what I've been doing.

Solve 4(cos^2(3x)) 3 = 0 for [0, 2π). Give exact answers.

Here is what I've done so far:

4 (cos^2(3x)) - 3 = 0

4 (cos^2(3x)) = 3 (added 3)

cos^2(3x) = 3/4 (divided by 4)


The three is throwing me off a little and I don't know what the next step is.

Please let me know if I can give anymore information or make anything clear.
You have \(\displaystyle cos^2(3x) = 3/4\). That square in the cosine is getting in the way, don't you think?

-Dan
 
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Apr 2012
72
0
Seattle, Washington
Thank you, Dan.

I determined:

4 (cos^2(3x)) - 3 = 0

4 (cos^2(3x)) = 3 (added 3)

cos^2(3x) = 3/4 (divided by 4)

√cos^2(3x) = √3/4

cos3x = +/- √3/(2)

3x = π/6 + (nπ)

and

3x = 5π/6 + nπ

Final answer (dividing by 3)

x = π/18 + (nπ)/3

and

x = (5π)/18 + (nπ)/3
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Since \(\displaystyle \displaystyle \begin{align*} \cos{(3x)} = \pm \frac{\sqrt{3}}{2} \end{align*}\), you should find that \(\displaystyle \displaystyle \begin{align*} 3x \end{align*}\) can give FOUR solutions (one for each quadrant of the unit circle), and then dividing by 3 will give you 12 solutions (as the period is diminished to \(\displaystyle \displaystyle \begin{align*} \frac{1}{3} \end{align*}\) of the original period).
 
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Apr 2012
72
0
Seattle, Washington
Hi Prove it,

I determined that there were the four solutions and then, dividing by three, I came up with:

x = π/18 + (nπ)/3

and

x = (5π)/18 + (nπ)/3


as my final answers.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
You have only gotten TWO of the possible solutions from the unit circle. There are FOUR.
 
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Apr 2012
72
0
Seattle, Washington
x = π/18 + (nπ)/3

and

x = (5π)/18 + (nπ)/3

and

x = (7π)/18 + (nπ)/3

and

x = (11π)/18 + (nπ)/3
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
That's better, although you should be adding \(\displaystyle \displaystyle \begin{align*} \frac{2n\pi}{3} \end{align*}\) to each, not \(\displaystyle \displaystyle \begin{align*} \frac{n\pi}{3} \end{align*}\).

Now that you have the four starting solutions, what are the twelve solutions in the region \(\displaystyle \displaystyle \begin{align*} x \in [0 , 2\pi ) \end{align*}\)?
 
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