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# Trigonometric Equations that Require Factoring - Problem 1

###### 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.

We are solving some harder trigonometric equations. Equations like this; 2 sine squared x minus 3 sine x plus 1 equals 0 require factoring and the way to do that is just to factor them as if they were quadratic equations. You’ll usually be given numbers that factor easily.

So for example you have a 2 sine x here and a sine x here. Those will multiply out to 2 sine squared x and then I need numbers that multiply out to 1. So question is plus or minus. In order to get a -3 I would need a minus sine x and a minus 2 sine x so these are both going to be negative and that tells me that the two equations I’m dealing with are sine x equals a half, that could make this equal to 0, or sine x equals 1. So I have to solve both of these equations.

Let’s take a look at the unit circle to see where sine x equals a half. Remember that sine x equals a half represents the points where the y coordinates on the unit circle are a half. So a point like this or this and you should memorize these points in the first quadrant, memorize the points that correspond to pi over 6 pi over 4 and pi over 3 this one happens to correspond to pi over 6, so the angle whose sine is a half it's in the first quadrant is pi over 6. So that’s one solution.

Also remember with sine equations that the supplement of a solution is also a solution. The supplement of this solution is 5 pi over 6 right it's pi minus pi over 6. 5 pi over 6 and that gives us the second solution, or 5 pi over 6, we use periodicity in the second.

Now I have sine of x equals 1. Well the only place for the sine is 1 is right here 0,1 let me use a different color. This angle is going to be pi over 2 and it is only the 1 because if you think about it the supplement of pi over 2 is also pi over 2. Pi minus pi over 2 is pi over 2. And so our solution here is x equals pi over 2. So because I had two different ordinary trig equations that came from this more complicated one, I’ve got 3 principle solutions let’s use periodicity for each one of them. So x equals pi over 6 plus 2nPi or pi over 2, plus 2n pi, or 5 pi 6 over 6 plus 2n pi.

Now remember by n I mean some kind of integer right n is an integer that allows for the possibility that n could be negative, n could be negative 5 for example pi over 6 plus 2 times negative 5 pi. Pi over 6 minus 10 pi. So you can take any even multiple pi an add it or subtract it to any of these answers and get a new answer. So this is our final answer.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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