Video Transcript
Find the solution set for 𝑥 given the cos of two 𝑥 plus 13 root three multiplied by the cos of 𝑥 is equal to negative 19, where 𝑥 exists between zero and two 𝜋.
The parentheses or curved brackets tell us that 𝑥 is strictly greater than zero and less than two 𝜋. This also indicates that we need to give our solutions in radians. We can solve the equation in this range by using our double-angle identities.
One of these identities states that the cos of two 𝑥 is equal to two cos squared 𝑥 minus one. We can therefore rewrite our equation as shown. We can then add 19 to both sides of the equation, giving us two cos squared 𝑥 plus 13 root three cos 𝑥 plus 18 is equal to zero. If we let 𝑦 equal cos of 𝑥, this can be rewritten as two 𝑦 squared plus 13 root three 𝑦 plus 18 equals zero.
We now have a quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, which we can solve using the quadratic formula as shown. Our values of 𝑎, 𝑏, and 𝑐 are two, 13 root three, and 18, respectively. Substituting these values into the formula, we get two values for 𝑦. Either 𝑦 is equal to negative root three over two or negative six root three. This means that the cos of 𝑥 must be equal to negative root three over two or negative six root three.
As the value of the cos of 𝑥 must lie between negative one and one inclusive, the second option has no solutions. We can now solve the equation the cos of 𝑥 is equal to negative root three over two using our CAST diagram. As the cos of our angle is negative, we will have solutions in the second and third quadrants. One of our solutions will be between 𝜋 over two and 𝜋 and our second between 𝜋 and three 𝜋 over two.
We know that the cos of 30 degrees or 𝜋 over six radians is equal to root three over two. This means that our two solutions will be equal to 𝜋 minus 𝜋 over six and 𝜋 plus 𝜋 over six. This gives us two solutions: five 𝜋 over six and seven 𝜋 over six. The solution set of the equation cos two 𝑥 plus 13 root three cos 𝑥 equals negative 19 are five 𝜋 over six and seven 𝜋 over six.
