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

Find the set of possible values of π‘₯ which satisfy 1/√(cosΒ²π‘₯ βˆ’ cos⁴π‘₯) = 2 where 0Β° < π‘₯ < 360Β°.

04:54

Video Transcript

Find the set of possible values of π‘₯ which satisfy one over the square root of cos squared π‘₯ minus cos to the fourth power of π‘₯ equals two, where π‘₯ is greater than zero degrees and less than 360 degrees.

In order to answer this question, we will begin by simplifying the left-hand side of our equation by using our knowledge of the Pythagorean and double-angle trigonometric identities. The denominator of the left-hand side is as shown. Factoring out cos squared π‘₯ from the expression under the square root gives us the square root of cos squared π‘₯ multiplied by one minus cos squared π‘₯.

One of the Pythagorean identities states that sin squared πœƒ plus cos squared πœƒ is equal to one. This can be rewritten as sin squared πœƒ is equal to one minus cos squared πœƒ. Using this, we can rewrite our expression as the square root of cos squared π‘₯ sin squared π‘₯. Our next step is to square root each part of the argument separately. Since we’re squaring cos π‘₯ and sin π‘₯ in the expression before taking the square root, either of them could be positive or negative and still give us a positive result when squared. This means that the square root of cos squared π‘₯ sin squared π‘₯ is the absolute value of cos π‘₯ sin π‘₯.

Next, we recall one of the double-angle identities: sin two πœƒ is equal to two sin πœƒ cos πœƒ. Dividing through by two, this can be rewritten as sin two πœƒ over two is equal to sin πœƒ cos πœƒ. And our expression can be rewritten as the absolute value of sin two π‘₯ over two. Taking the constant outside, the denominator of our equation is equal to a half multiplied by the absolute value of sin two π‘₯. Dividing one by this expression gives us two over the absolute value of sin two π‘₯. And from the initial equation, this is equal to two.

We can then solve our equation by firstly multiplying through by the absolute value of sin two π‘₯. Next, we divide through by two such that the absolute value of sin two π‘₯ equals one. This gives us two possible solutions: either sin two π‘₯ equals one or sin two π‘₯ equals negative one. We were told in the question that π‘₯ lies between zero and 360 degrees. This means that two π‘₯ must be greater than zero degrees and less than 720 degrees.

By sketching the graph of 𝑦 equals sin πœƒ between zero and 720 degrees, we can find the values of πœƒ for which sin πœƒ equals positive or negative one. There are two values of πœƒ, for which sin πœƒ equals one in this range. They are 90 and 450 degrees. In the same way, sin πœƒ equals negative one when πœƒ is equal to 270 and 630 degrees. We can therefore conclude that when sin two π‘₯ is equal to one, two π‘₯ is equal to 90 degrees and 450 degrees. Dividing through by two, this means that π‘₯ is equal to 45 degrees and 225 degrees. When sin two π‘₯ is equal to negative one, two π‘₯ is equal to 270 degrees and 630 degrees. Once again, we can divide through by two such that π‘₯ is equal to 135 degrees and 315 degrees.

There are four possible values of π‘₯ which satisfy the equation in the given range. They are 45 degrees, 135 degrees, 225 degrees, and 315 degrees.

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