uc3rmw_0505e04

Hi there. In this example, we will solve the following trigonometric equation, cosine of 2x equals 1/2. And we're going to solve that in the interval from 0 to 2 pi. The trigonometric equation contains a multiple angle-- that is, in which the argument of the trigonometric function is a multiple angle, in this case, 2x. Let's recall some examples of what equations with multiple angles are. We have the multiple angles highlighted in red in each of these three equations. We have sine of 3x, the multiply angle being 3x equals 1/2. Cosine of 1/2 x equals 0.7, the multiple angle here is 1/2 x. And of course 2x and 4x in the final equation. Step 1 is just to find the common angle for the trigonometric equation. We take the inverse cosine of the right-hand side of the equation. To find the common angle, inverse cosine of 1/2 equals pi over 3. Therefore, cosine of 2x is equal to the cosine of pi over 3. In step 2, we're going to go ahead and find all the solutions to the equation, cosine 2x equals cosine pi over 3, given by the table below. The relevant row in this table is the second row, which shows all the solutions for cosine equations. So in our case, we have that the solutions are that 2x, our multiple angle, is equal to pi over 3, the common angle, plus the periodicity 2n pi. And, 2x is equal to 2 pi minus pi over 3, plus our periodicity, again, 2n pi, which we can simplify to be 5 pi over 3 plus 2n pi. Dividing by 2 in both these solution sets, we have that x is equal to pi over 6 plus n pi, and x is equal to 5 pi over 6 plus n pi. We've divided 2 out throughout the solution set to actually isolate x and find the solutions for x. To find all the solutions in the interval from 0 to 2 pi, we're going to try different values of n. First, we're going to try n equals negative 1. And that's just on the first solution set. We find that with n equal minus 1, we have x equal pi over 6, minus pi, which reduces to minus 5 pi over 6. And next we'll try n equals 0 in this solution set. We have, with n equals 0, x is equal to pi over 6. And trying n equals 1, we have that x equals pi over 6 plus pi, which simplifies to 7 pi over 6. And for n equals 2, we find that x is equal to pi over 6 plus 2 pi. Finding the common denominator, this reduces to 13 pi over 6. Now we plug in the same values for the integer and for the second solution set. With n equal minus 1, we have that x equals 5 pi over 6, minus pi, which simplifies to minus pi over 6. Plugging in n equals 0, we just have that x equals 5 pi over 6. And plugging in that n equals 1, we have that x is equal to 5 pi over 6 plus pi, which reduces to 11 pi over 6. And finally, with n equal 2, x equals 5 pi over 6 plus 2 pi, which reduces to 17 pi over 6. So now we'll examine all the solutions that we found, of these various integer values of n, and find out which one of them lies in the interval from 0 to 2 pi. We find that the values resulting from n equals minus 1 are both negative and therefore outside the interval from 0 to 2 pi. And those resulting from n equals 2 are too large. They're outside the interval from 0 to 2 pi. Therefore, the solutions corresponding to n equals 0 and n equals 1 are those solutions that lie in the interval from 0 to pi. Those are pi over 6, 5 pi over 6, 7 pi over 6, and 11 pi over 6.