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Question Video: Solving a Trigonometric Equation Using Double-Angle Identities Mathematics • 10th Grade

Find the set of solutions in the range 0Β° < π‘₯ < 180Β° for the equation (sin π‘₯ + cos π‘₯)Β² = 2 sinΒ² 2π‘₯.

05:38

Video Transcript

Find the set of solutions in the range π‘₯ is greater than zero degrees and less than 180 degrees for the equation sin π‘₯ plus cos π‘₯ all squared is equal to two sin squared two π‘₯.

In order to answer this question, we will use our knowledge of the Pythagorean and double angle trigonometric identities. We will begin though by distributing the parentheses on the left-hand side of our equation. Using the FOIL method and then collecting like terms, we see that squaring sin π‘₯ plus cos π‘₯ gives us sin squared π‘₯ plus two sin π‘₯ cos π‘₯ plus cos squared π‘₯. The Pythagorean identity sin squared πœƒ plus cos squared πœƒ equals one means that we can rewrite sin squared π‘₯ plus cos squared π‘₯ as one.

We can also use the double angle identity sin two πœƒ is equal to two sin πœƒ cos πœƒ to rewrite the middle term of our expression. The left-hand side of the original equation can be rewritten as one plus sin two π‘₯. And this is equal to two sin squared two π‘₯. Subtracting sin two π‘₯ and one from both sides of our equation, we have two sin squared two π‘₯ minus sin two π‘₯ minus one is equal to zero. We now have a quadratic in sin two π‘₯. And we will let 𝑦 equal this.

Rewriting the equation in terms of 𝑦, we have two 𝑦 squared minus 𝑦 minus one is equal to zero. We can solve this quadratic by factoring, giving us two 𝑦 plus one multiplied by 𝑦 minus one is equal to zero. Since the product of the two parentheses equals zero, then at least one of two 𝑦 plus one and 𝑦 minus one must equal zero. We therefore have two solutions 𝑦 equals negative one-half and 𝑦 is equal to one.

Recalling our substitution, this means we’ll have solutions to the equation when sin two π‘₯ equals negative one-half and when sin two π‘₯ equals one. We were told in the question that π‘₯ is greater than zero degrees and less than 180 degrees. This means that two π‘₯ lies between zero and 360 degrees. By sketching the graph of 𝑦 equals sin πœƒ, we can find the values of πœƒ for which sin πœƒ equals negative one-half and one. The only value of πœƒ for which sin πœƒ equals one between zero and 360 degrees is 90 degrees. This means that if sin two π‘₯ is equal to one, two π‘₯ is equal to 90 degrees. Dividing through by two, we have our first solution π‘₯ is equal to 45 degrees.

Next, we see that the horizontal line 𝑦 is equal to negative one-half intersects the sin curve twice between zero and 360 degrees. One solution lies between 180 and 270 degrees, and the second lies between 270 and 360 degrees. Due to the symmetry of the sine function, these values are equal distance above 180 degrees and below 360 degrees. From our knowledge of special angles, we recall that sin of 30 degrees is equal to one-half. And due to the symmetry at either end of the sine graph, we see that the sin of 30 degrees will be equal to the negative sin of 330 degrees. This means that the sin of 330 degrees is equal to negative one-half. Adding 30 to 180 gives us 210. And using the symmetry once again, we see that the sin of 210 degrees is also equal to negative one-half.

This means that if sin two π‘₯ is equal to negative one-half, then two π‘₯ could be equal to 210 degrees or 330 degrees. We can then divide through by two once again, giving us solutions for π‘₯ equal to 105 and 165 degrees. We can therefore conclude that the set of solutions in the given range for the equation sin π‘₯ plus cos π‘₯ all squared is equal to two sin squared two π‘₯ is 45 degrees, 105 degrees, and 165 degrees.

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