Like what you saw?
Create FREE Account and:
- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics
Trigonometric Equations that Require Factoring - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
When solving trig equations that involve factoring here’s an example; tangent squared x minus tangent x minus 2 equals 0. Now this time to make it interesting I’m not look for solutions within the interval from 0 to pi that’s the meaning I want x to be the end the interval from 0 to 2 pi.
I still start the same way I want to factor this equation so I’m going to have two factors and because the first term is tangent squared, I’m going to start with tangent x on both of them. I just need to figure out what goes here and they have to be factors of 2, so one of them is going to be 1, one of them is going to be 2 probably.
Now let’s see what we need here in order to get a middle term of -10x, one of them has to be negative and this one has to be positive so this will work. This will multiply out to -2 and minus 2 10x plus 10x will give me -10x that works.
I think it’s a good idea to always check the factorization by mentally multiplying through because a lot of people don’t check and then they end up, if they had checked they wouldn’t have gotten the wrong middle term or something make sure that you double check these things.
Anyway this factorization gives me two equations I need to solve; tangent x equals -1 or tangent x equals 2. So I need to look for points where tangent is -1 or 2 so let’s take a look at the unit circle.
Now remember the tangent function is not like the sine and cosine function, tangent function will equal -1 where the terminals side of my angle has slope -1 so one such place will be negative pi over 4. Because this point will be root 2 over 2 negative root 2 over 2. And the slope is going to be rise over run negative root 2 over 2 divided by root 2 over 2 -1. Another point would be up here, we can take this negative pi over 4 and add pi to it to get 3 pi over 4 but that’s just using periodicity.
So x equals negative pi over 4 is our principle solution, remember the tangent equations only have one principle solution. We get the rest of them from periodicity. Now tangent x equals 2 it's going to turns out that this is not going to give us a specials value but just to draw a picture I want to draw the terminals side of my angle with slope 2.
So I’m just estimating let’s say that’s a slope 2. Unfortunately that does not correspond with any of our special values, I mean there is a special value with a tangent of root 3 with 1, with 1 over root 3, 2 is not one the specials values so I’m going to have to actually use inverse tangent here.
It’s important to know that when you do use inverse tangent, so x equals inverse tangent 2, what you get is the principle solution between negative pi over 2 and pi over 2. So I will get the answer. This answer will be inverse tangent of 2.
Now if your teacher wants an exact answer this is the way to leave it, if they want an approximate answer a decimal answer you can use your calculator and there will be something like 1.107 anyway so these are my two principle solutions. So remember after that you are going to use periodicity to get the rest of course I have to keep in mind I don’t need all the solutions but I want the ones that fit in this interval, so let me start adding pi to each of this and see how many I get that are actually in the interval.
This one as you can see are not in the interval, this is a little to the left of 0, so I need to add pi and get 3 pi over 4. That one is in the interval. If I add another pi I get 7 pi over 4. That’s also an interval so let me just record over here x equals 3 pi over 4, 7 pi over 4 those are solutions. And then this number 1.107 is definitely in the interval of 0 to 2 pi. 2 pi is like 6.28 so this number is definitely between 0 and 6.28.
I could add pi to get the next one, inverse tangent 2 plus pi but I can’t add another pi because I’m going to have something bigger than 6.28. So whatever value this is I’m just going to write the final solution is or x equals inverse tangent of 2, inverse tangent of 2 plus pi. Those would be my solutions.
Just remember whenever you have to deal with a number that doesn’t represent a special angle, don’t be afraid to write this as your final answer If your teacher wants an exact value, but if they want approximations you got to use your calculator for these non-special values. So again inverse tangent of 2, 1.107 if I added pi 4.249. So this would be approximately 1.107, 4.249.
Please enter your name.
Are you sure you want to delete this comment?
Sample Problems (3)
Need help with a problem?
Watch expert teachers solve similar problems.