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In this example, we will solve the following multiple-angle trigonometric equation-- sine of x/2 equals minus square root of 3/2 in the interval from 0 to 2 pi. The trigonometric equation contains a multiple angle. That is, the argument of the trigonometric function is a multiple angle-- in this case, x over 2. The first thing to do when solving a trigonometric equation is to isolate the trigonometric function on the left-hand side of the equation. In our case, this is already done, so we're going to go ahead and move on to step 1 here and find the common angle, alpha. The common angle is found by taking the inverse sine of the right-hand side of the equation. That is, take the inverse sine of minus the square root of 3 divided by 2. The common angle is minus pi over 3. So we have that the equation that we want to solve is sine of x over 2 equal to sine of minus pi over 3. Now we're going to move on to step 2, and we want to find all the solutions of the equation sine of x over 2 equals sine of minus pi over 3. We recall that for the sine function, the table shows that all of the solutions to the sine function are given by x equal alpha plus 2n pi, alpha being the common angle-- or x equal pi minus alpha plus 2n pi, 2n pi being the periodicity of the sine function. Thus, the solutions to our equation are x/2 equal to the common angle-- minus pi over 3-- plus 2n pi, and x/2 equal to pi minus minus pi over 3 in parentheses plus 2n pi, which simplifies to 4 pi over 3 plus 2n pi. But we're not quite done. We need to multiply both sides of each of these solution sets by 2 in order to isolate x. Multiplying each side by 2, we have for the first solution set x equals minus 2 pi over 3 plus 4n pi, and x equals 8 pi over 3 plus 4n pi. Now we want to move on to step 3. We find that for any integer n, the values of x in both expressions in step 2 lie outside the interval from 0 to 2 pi. For example, when n equals 0, the solution set on the right gives x equals minus 2 pi over 3, which is less than 0 and outside our interval from 0 to 2 pi. And for n equals 1, we have that x equals minus 2 pi over 3 plus 4 pi, which is 10 pi over 3. And that solution is greater than 2 pi, lying outside of the interval from 0 to 2 pi. Therefore, the equation sine of x/2 equals minus root 3/2 has no solution in the interval from 0 to 2 pi. Therefore, the solution is the null set.

Video Details

Duration: 3 minutes and 10 seconds
Country: United States
Language: English
License: Dotsub - Standard License
Genre: None
Views: 12
Posted by: 3play on Jul 24, 2017

Please translate to spa_la. Account ID: 585. Notes on format and other things are here: http://s3.amazonaws.com/originp3/app/translation-profiles/profiles/c728d56a6e3afc44c0a63b925c143995.html

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