# Thread: Solving trigonometric linear equation

1. ## Solving trigonometric linear equation

Hi,

I hope someone can help. I would like some clarification on solving trigonometric functions. Let me provide an example:

Is there a way to directly solve for 3t on my calculator? From what it seems, you have to first solve as if the equation is cost = 1/2 and then (assuming that the domain is between 0 and 2pi), we find the two corresponding related acute angles. Once we have these related acute angles, we can then sub them in for t within the argument 3t = 1/2 and then isolate t. This seems like a rather long process, and I'm curious if there is any easier way to calculate for t?

2. ## Re: Solving trigonometric linear equation

$\cos (3t) = \dfrac{1}{2} \Longrightarrow 3t = \arccos\left(\dfrac{1}{2}\right) \Longrightarrow t = \dfrac{1}{3}\arccos\left(\dfrac{1}{2}\right)$

This gives you a way to plug it directly into your calculator. But, the $\arccos$ function only gives a single result. It does not give you all angles. It only gives you angles in the range 0 to pi. So, you will still need to find the rest of the angles.

3. ## Re: Solving trigonometric linear equation

You can also use Wolframalpha:

https://www.wolframalpha.com/input/?i=cos(3t)%3D1%2F2

It tells you a general formula for finding values of $t$. There are six possible values for $t$:

$\dfrac{\pi}{9}, \dfrac{5\pi}{9}, \dfrac{7\pi}{9}, \dfrac{11\pi}{9}, \dfrac{13\pi}{8}, \dfrac{17\pi}{9}$

4. ## Re: Solving trigonometric linear equation

Alright so by computing 1/3arccos(1/2), I get pi/9. This would be my first angle.

How would I calculate the second angle that is also between 0 and 2pi? I know that the second angle would be in quadrant 4, but I'm sure entirely sure how to find that angle. I tried subtracting pi/9 from 2pi, and I didn't get the correct answer.

I also tried finding the solutions between 0 and 2pi for the equation cos(pi/6)t=-2/3.5. I calculated pi/6arccos(2/3.5) to the value of t, and then I got the first solution by subtracting t from pi and then got the second solution by adding pi to t. Do you see where I went wrong in these steps?

5. ## Re: Solving trigonometric linear equation

Originally Posted by otownsend
Alright so by computing 1/3arccos(1/2), I get pi/9. This would be my first angle.

How would I calculate the second angle that is also between 0 and 2pi? I know that the second angle would be in quadrant 4, but I'm sure entirely sure how to find that angle. I tried subtracting pi/9 from 2pi, and I didn't get the correct answer.

I also tried finding the solutions between 0 and 2pi for the equation cos(pi/6)t=-2/3.5. I calculated pi/6arccos(2/3.5) to the value of t, and then I got the first solution by subtracting t from pi and then got the second solution by adding pi to t. Do you see where I went wrong in these steps?
I offered a second way of solving the problem, but it is not necessarily easier. I think the easiest approach for you would be to go back to the original method. Find angles for $\theta$ so that $\cos \theta = \dfrac{1}{2}$. Then, you have $3t = \theta$, so $t = \dfrac{\theta}{3}$.

6. ## Re: Solving trigonometric linear equation

For $0 \leq t \leq 2 \pi$, this is equivalent to $0 \leq 3t \leq 6 \pi$. Look for solutions for $3t$ in that range, then divide by $3$.

So you get $$3t= \frac{\pi}{3}, 2\pi - \frac{\pi}{3}, 2\pi+ \frac{\pi}{3},..., 6\pi - \frac{\pi}{3}=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3},...,\frac{17\pi}{3}$$

Then dividing by 3 gives all the solutions (given by SlipEternal).

Have you seen a CAST diagram? If you haven't, I highly recommend you get used to using it, it makes these types of problems a lot easier.

7. ## Re: Solving trigonometric linear equation

unit circle with special angles annotated

learn it, live it, ... love it.

8. ## Re: Solving trigonometric linear equation

Thanks for all your help. I understand now.