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# Solving Trigonometric Equations - 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 talking about trigonometric equations I have an example here, cosine of x equals minus root 2 over 2. Now remember that a trigonometric equation is going to have infinitely many solutions. I’ve got the cosine graph drawn here and I’ve got y equals negative root 2 over 2 drawn on the graph and everyone of these points where the two curves intercept is going to represent our solutions specifically I’m looking for the x coordinates.

Now I don’t usually solve a trig equation on the graph like this, usually I go to the unit circle. I just want to make the point that they are infinitely many solutions and that there are two solutions per period. So let’s go to the unit circle.

Now one of the solutions is represented by this angle. Now how do we find that angle? Remember that the way sine and cosine are defined they are defined as x and y coordinates of this point at the terminal side of this angle. So I’m going to put coordinates negative root 2 over 2 something. I need to figure out what angle has a cosine of negative root 2 over 2, so I could use x equals inverse cosine of negative root 2 over 2. But you also remember that the points on the unit circle with coordinate roots 2 over 2 and root 2 over 2, these are the multiples of 45 degrees of pi over 4. And so this angle is probably pi over 4 and this angle will then be 3 pi over 4.

Remember that inverse cosine is going to give you an angle between 0 and pi and that’s all it's going to give you, it’s not going to give you all the solutions to your equation.

To get another solution I have to use the fact that cosine is isn’t even function that’s what this equation means. If you take opposite inputs and out them in cosine you get the same output. The two outputs are exactly the same and you kind of see that on the unit circle, this angle is the opposite angle. This is 3 pi over 4, this is -3 pi over 4, they have the same x coordinate at these two points and that means that the two angles have the same cosine value, so the cosine of -3 pi over 4 is also negative root 2 over 2 and that means x equals -3 pi over 4 is another solution.

Now remember I said there are two solutions per period I’ve just found two solutions in one period, to find the rest of the solutions I have to use periodicity. The fact that cosine is a periodic function with period 2 pi. That means that I can add any integer multiple of 2pi to either of these solutions and get the rest of them. So the solutions are 3pi over 4 plus 2 n pi or -3pi over 4 plus 2nPi. Or you could write all in one step x equals plus or minus 3 pi over 4 plus 2nPi.

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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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