###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Solving Trigonometric Equations - Problem 3

Norm Prokup
###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Let’s talk about another trig equation, this time a secant equation. Now a secant equation involves the reciprocal function secant. Anytime you see a reciprocal function, you want to turn this into a trig equation involving sine cosine or tangent.

So the first thing I’m going to do is call this 1 over cosine of 3x equals 2, because secant equals 1 over cosine. And then I can take the reciprocal of both sides and I get cosine of 3 equals a half. And so you see that I’m in two steps I know that I no longer have to deal with the secant function and this is always true you never have to deal with the reciprocal trig functions when you are solving trig equations. You can go back to one of the familiar ones.

So we have cosine of 3x equals one half, let me make a little substitution here. This is what I usually do when I have something other than just the variable inside my trig function, I’ll call this theta. And so my equation becomes cosine of theta equals a half. And I can find the solutions to this equation on the unit circle. So here is the unit circle, I need to find angles that make the first coordinate of this point one half, because that’s how cosine is defined.

So if you recall pi over 3 does that. Of course this problem asks us to solve for x in the interval from 0 degrees to 360 degrees. This tells us two things, first of all that we are not going to want infinitely many solutions and second that we are working in degrees and not radian. So let me immediately switch to 60 degrees, pi over 3 is 60 degrees, that gives me 1 solution.

How to get another? Remember that sine and cosine each have two solutions usually per period. There’s going to be another solution and with cosine the other solutions due to the fact that cosines are even function, opposite inputs have the same output. You can see that on the unit circle by thinking what are the points on the units circle has an x coordinate of one half? There’s one down here and it's going to be the mirror image of this point. So you are going to have one half common negative y whatever the y coordinate is and see you are going to have the second solution being -60 degrees by symmetry. So for -60 degrees.

Now to get the remainder of the solution we have to use periodicity so we’ve got theta equals plus or minus 60 degrees and when you are dealing with degrees, periodicity means you can add an integer multiple of 360 degrees. Normally we add integer multiple of 2 pi, because 2 pi is one radian but then degrees are one revolution rather. In degrees one revolution is 360 degrees.

So we are about ready to re-substitute, remember that I substituted theta for 3x. So we are going back to 3x and that’s what we get. All we have to do is divide by 3. So dividing by 3 gives me plus or minus 20 degrees plus n times 120.

Now the problem asked me to find only solutions between 0 and 360, so I don’t want to give this as my final answer because I'm not exactly reading the instructions here, so let me start with 20 degrees. Plus 20 degrees plus 0 times 120 that’s the smallest solution I’m going to get. And then let me start with the 20 degrees and keep adding integer multiples until I leave the interval from 0 to 360. So I’m going to add 120 and get 140.

That one is good. I add 120 again and get 260, also good. Add 120 again and get 380 not the interval, so I wouldn’t include that. Now I know that -20 is not a solution but when I add multiples of 120 degrees I will get solutions, so let me add 120 and get 100 degrees that’s the solution, that one will be. Add another 120, 220, add another, 340 and you can see that if you add another 120 you’ll be outside of the interval. So these are your solutions. 20 ,140, 260, 100, 220 and 340.

And notice the way I worked this out I started with the positive 20 and I added multiples of 120. Here I started with -20 and I added multiples of -120 and I'll just chose this as my final answer so the values actually fall in the intervals 0 and 360, that’s your final answer.
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