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Trigonometric Equations that Require Factoring - Concept

Teacher/Instructor 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.

Solving second degree trig functions can be accomplished by factoring polynomials into products of binomials. When factoring trigonometric equations, we can use the zero product property to set up two first degree trig equations that you can solve using the unit circle. If an equation has sine and cosine, we substitute for one with an identity.

I want to solve some trigonometric equations that require factoring. You may see some examples like this 4sine squared theta equals 2, 2sine squared x minus 3sine x plus 1 equals 0 and so on. These kinds of trig equations they look really complicated but they're not, they're they're basically just simple trig equations kind of disguised algebraically.
Let's start with an example let's do the first one 4sine squared theta equals 2. Like I said these these equations require factoring and the reason we factor is to reveal what the the simpler trig equations that are hiding inside are. We have 4sine squared theta minus 2 equals 0 and we can factor that as a difference of squares, 2sine theta minus root 2, 2sine theta plus root2 and so that gives us two results. First sine theta could equal to 2 over 2 or from this factor sine theta could equal negative root 2 over 2 and each of this is just so simple sine equation so we have to solve these two guys.
Let's take a look at the unit circle. Now we write my equations here I've got sine theta is equal to 2 over 2 sine theta equals negative root 2 over 2. Let's focus on the first one, for the sine of theta to be root 2 over 2 that means I have to find a point on the unit circle whose y coordinate is root 2 over 2 so here something root 2 over 2 and here. Now when I look at these points I remember that there are lots of points that have coordinates plus or minus root 2 over 2 and they occur at integer multiples of 45 degrees or pi over 4 so this this angle is pi over 4 and here's another one, this angle is 3pi over 4 and again both of these points have a y coordinate of root 2 over 2. Remember when you're solving a trigonometric equation a sine equation particulaly, if you find one solution another solution can be found by taking the supplement because the sine of two supplementary angles will always be the same so another solution will be pi minus pi over 4 which is where we get the 3 pi over 4 so that gives us two principle solutions and if you want to get the rest of them, you just add 2n pi right from periodicity. I can take pi over 4 and add to 2 pi to it that would still be a solution because that would give me the same point on the unit circle.
I could add another 2 pi or I subtract any integer multiple of 2 pi or 3 pi over 4 plus 2n pi. Now what about the second equation? Sine theta equals negative root 2 over 2. Well negative root 2 over 2 would be the reflection of this point across the x axis, x minus root 2 over 2 and so here here is my angle and it makes sense that this would be the reflection of this angle negative pi over 4 and again if you if you have one solution negative pi over 4, another solution will be the supplement so pi minus negative pi over 4 and of course that's pi minus negative 5 or, that's 5 pi over 4, so these are two principle solutions and incidentally 5 pi over 4 is this angle starting here going around to there that's 5 pi over 4 you might say wait a second, isn't that also negative 3 pi over 4? And you can call it either one so in our final solutions we have to use periodicity so we get negative pi root 4 plus 2pi 2n pi or 5 pi root 4 plus 2n pi, so remember when you're giving your final answer all of these are solutions, you want to put a big or in between. All of these are possibilities you could have pi over 4 plus 2n pi, 3 pi over 4 plus 2n pi, -pi over 4 plus 2n pi or 5 pi over 4 plus 2n pi.

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