Basic Trigonometric equation

We’re now going to continue our discussion of the equations by looking at trigonometric equations, particularly simple-looking trigonometric equations. However, we’ll see that there are definitely methods to know and pitfalls to avoid, even for these simple equations. Let’s begin with sine x equals alpha. So alpha’s a given number. To begin with, we can say that if the absolute value of alpha is bigger than 1, then there are no solutions. Why? Because sine x is always between minus 1 and plus 1. So there can’t be any x’s that satisfy the equation. If, on the other hand, the absolute value of alpha is less than 1, then there is a number a satisfying sine of a equals alpha.

In fact, there is one that we’ve identified– a equals the arcsine of alpha. That’s the a that lies in the interval minus pi over 2 to pi over 2. Let’s see how this works on the circular definition of our trigonometric functions. We measure out the angle a, and we see that the y-coordinate of the corresponding point on the unit circle is sine a. Are there any other numbers that would give you the same sine as a? Well, obviously, you could take a and add, let’s say, a positive integer multiple of 2 pi. That would bring you back to the same point on the unit circle, and hence, the same sign.

Or you could add a negative integer multiple of 2 pi, and you would still come back to the same point. So therefore, our solution set to sine x equals alpha will include not just a, but a plus 2 k pi, where k is any positive or negative integer. That is, an element of what we called z. And is that it? Does that exhaust all the possibilities? Well, it doesn’t, because you see that if you take the point to the angle a, but now you go out from the point on the unit circle to the one that corresponds to it in the left toward direction horizontally, you reach another point that has the same y-coordinate and, hence, the same sine.

What is that angle that gives you that point? Well, a moment’s thought will show you that the angle is the angle pi minus a. That’s what we call the supplementary angle to a. Remember two angles are supplementary if their sum is equal pi, and in that case, they have the same sine. We’ve seen this before with our trigonometric identities. So we need to take account of that supplementary angle pi minus a. It’ll give you the same sine, and also, of course, we can add to it any negative or positive integer multiple of 2 pi. And again, the sine will be the same. So to summarize, here’s the set of solutions to sine x equals sine a.

It’s either a, or the supplementary pi minus a, or else an angle that differs from one of those by an integer multiple of 2 pi. Let’s see how to use this to solve the following trigonometric equation. We have the sine of something involving x equals the sine of x. Given the previous discussion, you can see that the angle on the left, 2 x plus pi over 6, has to be either x itself or the supplementary angle to x, the angle pi minus x. Or something that differs from a, of course, an integer multiple of 2 pi from one of those. So these are the possibilities.

Now we can clean that up a little bit by putting all the x’s on the left-hand side. And we can simplify the expression a little further, and we come to the set of all possible x’s solving our equation. Let’s now turn to the equation cosine x equals alpha. Once again, this will only have solutions if the absolute value of alpha is less than or equal 1. And in that case, there’s an angle a that has cosine equals alpha, namely the angle we called arccosine or inverse cosine of alpha. And it’s in the interval 0 pi. So our equation then becomes cosine x equals cosine alpha.

Now here, instead of this supplementary formula that figured in the case of sine, it’s the evenness of cosine that will provide the other set of solutions other than those that are essentially a modulo, an integer multiple of 2 pi. So to summarize, cos x equals cos a will have the following solutions. Either a, or minus a, or an angle that differs from one of those by an integer multiple of 2 pi.

Let me point out a pitfall that can arise in solving these kinds of equations where you don’t add the period right at the beginning but only at the end. And that’s a slightly lazy way to do it, and it can lead to wrong answers. Consider this simple equation. Cosine of 3 x equals 1/2. Well, based on our previous discussion, we could perhaps solve it in the following incorrect way. We could say the 3 x has to either be the angle whose cosine is 1/2 and is between 0 and pi. That’s pi over 6. Or it could be minus pi over 6. That’s how we solve these cosine equations. Notice we haven’t added the 2 pi business yet.

Therefore, we find x equals plus or minus pi over 18. And now we add the 2 k pi, and we come to this set of answers. Let’s contrast that with the correct solution. The correct solution would start out the same way, except it would immediately add the possibility of differing from plus or minus pi over 6 by an integer multiple of 2 pi. Then when we simplify the answer, we find something which is different from the answer on the left. How is it different? Well, it allows more possibilities than there were in the ones we found on the left. That is, the first incorrect method failed to capture all the solutions.

And finally, for our simple trigonometric equations, let’s look at tangent x equals alpha. That’s going to be easier. Why? Because there’s a unique solution to tangent x equals alpha. For any alpha, whatever, there’s a unique solution in the interval minus pi over 2 to pi over 2. That’s how we define the arctan function. So one solution is arctan alpha. We also know that the tangent function is periodic of period pi. So to find all the other solutions, you simply have to take this first initial solution. And you add to it any integer multiple of pi, where k the integer is either positive or negative. And that will be the full set of solutions.