Trigonometric Equations find all solutions

[1MileCrash]

Homework Statement

sec 3\Theta = 2

Find all solutions within 0 < \Theta < 2\Pi

Homework Equations

none

The Attempt at a Solution

I am having trouble "knowing" what angles satisfy sec = 2 within the interval.

I can find one... ArcCos 1/2 = 1.047... which I can guess is \Pi/3.

The other is 4\Pi/3, but I have NO idea how to get that angle.

I need both to format two different general formulas to get all possible answers, but I have no problem doing that. My problem is the first step, magically knowing the angles that satisfy sec = 2 within the interval. I can find one using inverse functions but I don't know how one comes to the second.

 

[eumyang]

Is this the equation?
\sec 3\theta = 2

If so, neither π/3 or 4π/3 are solutions. (Plug them into the equation and see.) I suspect that you're thinking that if the secant of (some angle) = 2, then (some angle) = π/3. But (some angle) is actually 3θ, so 3θ = π/3. What is theta, then?

If you are looking for angle with a specific trig ratio, and you are sure that the angle is one of those special angles (based on 30°, 45°, 60°, ...), then draw triangles in the cartesian plane. And that's triangles, plural, because you should know that the secant of an angle is positive in two quadrants, right? You already got one: sec π/3 = 2. But where else?

After getting the 2nd angle, you will have to contend with the restriction of 0 ≤ θ < 2π. Remember, the angles we are finding equal 3θ, not θ. So given the restriction, you can say that 0 ≤ 3θ < 6π. You'll need to get some coterminal angles between 0 and 6π. (Hint: there will be six answers total.)

 

[1MileCrash]

I know that pi/3 and 4pi/3 aren't solutions to the overall equation, they are solutions to sec theta= 2. I use those angles to find the solutions. I just want a faster way to get the initial angles needed (that satisfy secant = 2 in this example.)

I also get 6 answers, pi/9; 7pi/9, 13pi/9, 4pi/9, 10pi/9, 16pi/9. These are easily found once I know the second angle that satisfies secant of 2. I may have overcomplicated the question. I just need a quick way of finding that second initial angle. The first one I find with arccos .5, the second one I usually have to draw a plane for which I don't have time for during exams.

Sorry I'm not using proper formatting, doesn't seem to work on my browser well.

 

[The Chaz]

sec(3t) = 2
invert (flip) both sides
cos(3t) = 1/2
Get a unit circle. Each point is of the form (cosine(angle), sine(angle))
We want the angles where cosine is 1/2.
These angles are:
π/3 and 5π/3
But since we have 3t, we include all solutions less than 6π. See the above reply for more details.
So 3t = π*(1/3, 5/3, 7/3, 11/3, 13/3, 17/3)
then t = (those, but divided by 3)

 

[eumyang]

I know that pi/3 and 4pi/3 aren't solutions to the overall equation, they are solutions to sec theta= 2. I use those angles to find the solutions. I just want a faster way to get the initial angles needed (that satisfy secant = 2 in this example.)

I also get 6 answers, pi/9; 7pi/9, 13pi/9, 4pi/9, 10pi/9, 16pi/9. These are easily found once I know the second angle that satisfies secant of 2. I may have overcomplicated the question. I just need a quick way of finding that second initial angle. The first one I find with arccos .5, the second one I usually have to draw a plane for which I don't have time for during exams.

Sorry I'm not using proper formatting, doesn't seem to work on my browser well.

As The Chaz stated, the 2nd angle wasn't 4π/3, but 5π/3. cos 4π/3 = negative 1/2.

If you're looking for a quick way of finding the angles in the various quadrants, I can tell you that when I learned trig I memorized the following:
- the sign of the trig functions in each quadrant
- the operations to perform to find angles in all quadrants given a reference angle

Reference angles are between 0 and π/2. Given a reference angle t, the corresponding angles in the 4 quadrants are as follows.
- Quadrant I: angle is just t.
- Quadrant II: angle is π - t.
- Quadrant III: angle is π + t.
- Quadrant IV: angle is 2π - t.

So in the case of cos θ = 1/2, you know that θ = π/3 if θ is in the 1st quadrant. Cosine is also positive in the 4th quadrant, so our corresponding angle would be
2π - π/3 = 5π/3.