# Math Help - Trigonometric equation

1. Yes I already checked, the answer is x=43°, x=403°... would there be two more answers like you said, or were you just talking about the results before checking for extraneous?

2. There are two more solutions. Just not 223 or 583. What assumption did you make that got you those two numbers?

3. You really need be familiar with trig functions of special angles. The ratio $\frac{\sqrt{3}}{2}$ can be found from the 30-60-90 special triangle. In our case,
$cos 30^{o} = \frac{\sqrt{3}}{2}$.
Also know the cartesian plane. Cosine is positive in Quadrants I and IV, so you can find the angle related to 30 degrees in Quadrant IV by subtracting it from 360 degrees. So
$cos 330^{o} = \frac{\sqrt{3}}{2}$
is also true.

It looks like you're already familiar with coterminal angles, so between that and what I have above, you should be able to find the four solutions.

4. I'm still missing something. I tried adding 90 degrees instead of 180, and that did not work. 2cos(43+90-13)=-1, and 2cos(43+270-13)=1.

5. What's missing is an understanding of the symmetry of the cosine function. You can flip it about a vertical axis of $x=n\pi$, and its graph will look the same. $2\pi$ is the smallest number $y$ such that $\cos(x)=\cos(x+y)$ for all $x\in\mathbb{R}.$ So trying to find solutions based on any smaller periodicity of the cosine function isn't going to work, I'm afraid.

So, you've got one answer you know is right: 43 degrees. Flip that about the $\pi$ axis. What do you get?

6. Originally Posted by wiseguy
I'm still missing something. I tried adding 90 degrees instead of 180, and that did not work. 2cos(43+90-13)=-1, and 2cos(43+270-13)=1.
I don't know why you are adding 90 degrees.

$cos 30^{o} = \frac{\sqrt{3}}{2}$
So if
$cos (x - 13^{o}) = \frac{\sqrt{3}}{2}$
then 30° = x - 13°, or x = 43°, which you got.

But
$cos 330^{o} = \frac{\sqrt{3}}{2}$ as well.
So what is x in this case?

7. x=317... I'm baffled at how I should show work for this, even though it seems relatively simple. Maybe I'll figure it in a few minutes after I post this

8. Maybe you can draw triangles in the 1st and 4th quadrants? sqrt(3) over 2 is a very common trig ratio, so if I were to be grading this problem I probably would accept work like this:

\begin{aligned}
cos (x - 13^{o}) &= \frac{\sqrt{3}}{2} \\
cos (x - 13^{o}) &= cos 30^{o}\,&\text{OR}\,cos (x - 13^{o}) &= cos 330^{o} \\
x - 13^{o} &= 30^{o}\,&\text{OR}\,x - 13^{o} &= 330^{o} \\
\end{aligned}

... and so on.

BTW, it's not x = 317. Add 13 degrees, not subtract.

9. Where do the first and forth quadrants thing come from?

10. I'm aware of the cartesian plane that you mentioned before, however I cannot find it in my textbook.

11. Cosine is positive in the first and fourth quadrants.

Have you learned about reference angles? Take a look at this page from Spark Notes:
SparkNotes: Trigonometry: Trigonometric Functions: Reference Angles

I know that $cos 30^{o} = \frac{\sqrt{3}}{2}$, and that cosine is also positive in the 4th quadrant. To find the angle whose reference angle is 30°, subtract 30° from 360° (this is true for any angle in the 4th quadrant) and you get 330°. So $cos 330^{o} = \frac{\sqrt{3}}{2}$ is also true.

12. cos(x-13°)=√3/2
x-13°=arccos√3/2
x-13°=0.5235987756
x=13°+30°
x=43°

360°-30°=330°

x=13°+330°
x=343°

How do I show that x also equals 403 and the missing forth value using this same method?

13. Originally Posted by wiseguy
How do I show that x also equals 403 and the missing forth value using this same method?
Cosine has a period of 2π or 360°, so cos (x + 360°) = cos x. In other words, add 360° to the two answers you got above.

14. I got 43, 343, 403, and 703... how does it look? Thanks so much this :censored: thing has been driving me nuts!

15. Originally Posted by wiseguy
I got 43, 343, 403, and 703... how does it look? Thanks so much this :censored: thing has been driving me nuts!
Looks good! For sure, it's alot to keep track of when dealing with trig. I know, trig made my head spin the first time I tried to learn it. Just keep at it!

Page 2 of 2 First 12