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

Find the set of possible values of 𝑥 which satisfy 1/√(cos²𝑥 − cos⁴𝑥) = 2 where 0° < 𝑥 < 360°.

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### Video Transcript

Find the set of possible values of 𝑥 which satisfy one over the square root of cos squared 𝑥 minus cos to the fourth power of 𝑥 equals two, where 𝑥 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 𝑥 from the expression under the square root gives us the square root of cos squared 𝑥 multiplied by one minus cos squared 𝑥.

One of the Pythagorean identities states that sin squared 𝜃 plus cos squared 𝜃 is equal to one. This can be rewritten as sin squared 𝜃 is equal to one minus cos squared 𝜃. Using this, we can rewrite our expression as the square root of cos squared 𝑥 sin squared 𝑥. Our next step is to square root each part of the argument separately. Since we’re squaring cos 𝑥 and sin 𝑥 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 𝑥 sin squared 𝑥 is the absolute value of cos 𝑥 sin 𝑥.

Next, we recall one of the double-angle identities: sin two 𝜃 is equal to two sin 𝜃 cos 𝜃. Dividing through by two, this can be rewritten as sin two 𝜃 over two is equal to sin 𝜃 cos 𝜃. And our expression can be rewritten as the absolute value of sin two 𝑥 over two. Taking the constant outside, the denominator of our equation is equal to a half multiplied by the absolute value of sin two 𝑥. Dividing one by this expression gives us two over the absolute value of sin two 𝑥. 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 𝑥. Next, we divide through by two such that the absolute value of sin two 𝑥 equals one. This gives us two possible solutions: either sin two 𝑥 equals one or sin two 𝑥 equals negative one. We were told in the question that 𝑥 lies between zero and 360 degrees. This means that two 𝑥 must be greater than zero degrees and less than 720 degrees.

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

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