# Video: Solving Trigonometric Equations with the Cosine Function

Find the set of values satisfying 6 cos² 𝜃 − 7 cos 𝜃 − 5 = 0 where 0° ≤ 𝜃 < 360°. Give your answers to the nearest minute.

03:50

### Video Transcript

Find the set of values satisfying six cos squared 𝜃 minus seven cos 𝜃 minus five equals zero, where 𝜃 is greater than or equal to zero and less than or equal to 360 degrees. Give your answers to the nearest minute.

Notice that we have a trigonometric equation and that this trigonometric equation looks a little bit like a quadratic. In fact, we’re going to treat it a lot like one. We’re going to begin by defining 𝑥 to be cos 𝜃. And replacing cos 𝜃 with 𝑥 in our original trigonometric equation and we find that six 𝑥 squared minus seven 𝑥 minus five must be equal to zero. We’re going to begin then by solving this equation for 𝑥.

And in fact, the expression six 𝑥 squared minus seven 𝑥 minus five is fully factorable. We know it can be written as the product of two polynomials. The first term in each polynomial must have a product of six 𝑥 squared, whereas the second term must have a product of negative five. Then, when we multiply the outer terms and inner terms and find their sum, we need the solution to be negative seven 𝑥. The only way for these criteria to be satisfied is if our polynomials are three 𝑥 minus five and two 𝑥 plus one.

Now, the product of these two polynomials is equal to zero. And the only way for this to be true is if either three 𝑥 minus five is equal to zero or two 𝑥 plus one is equal to zero. We solve this first equation by adding five to both sides and then we divide through by three. And we see that 𝑥 is equal to five-thirds. Similarly, we solve the second equation for 𝑥 by subtracting one from both sides and then dividing by two. And we see that 𝑥 here is equal to negative one-half. Now, we originally said that 𝑥 was equal to cos 𝜃. So we can now say that cos 𝜃 must be equal to five-thirds or negative one-half.

We’re going to look at solving each of these equations for 𝜃. However, we know the absolute maximum value of cos 𝜃 to be one. So this means there’s no value of 𝜃 such that cos 𝜃 is equal to five-thirds. And here, we say 𝜃 is undefined. We can solve the second equation though by finding the inverse cos of both sides. So 𝜃 is equal to the inverse cos of negative one-half. That gives us that 𝜃 is equal to 120 degrees. And so, we have found one value of 𝜃, which satisfies this equation. However, it’s not the only one. And there’re a couple of different ways that we can find the second.

One method involves sketching the graph of 𝑦 equals cos 𝜃. We’re solving the equation cos 𝜃 equals negative one-half. So we add the line 𝑦 equals negative one-half to our diagram. We saw that the first solution, the first value of 𝜃 for which this is the case, is 120 degrees. So that’s the first point of intersection between the graph of cos 𝜃 and negative one-half. The second solution is over here. Now, in fact, we can use the fact that the graph is completely symmetrical about the line 𝜃 equals 180 degrees to find this value. We subtract 120 degrees from 360. And then, we find the other value of 𝜃 in the interval 𝜃 is greater than or equal to zero and less than or equal to 360 is 240 degrees.

The other method you might be confident with is the cos diagram. We are solving the equation cos 𝜃 equals negative one-half. So we used our cos diagram to identify the fact that cos is positive here and here. This, therefore, means it’s negative here and here. And so, our first solution for 𝜃 lies in the second quadrant. We see that our second solution lies in the third quadrant. We know the angles around a point sum to 360 degrees. So we can find our second solution once again by subtracting 120 from 360 to get 240 degrees.

And so, the set of values satisfying the equation six cos squared 𝜃 minus seven cos 𝜃 minus five equals zero, where 𝜃 is greater than or equal to zero and less than or equal to 360 degrees are 120 degrees and 240 degrees.