SOLUTION: Write the equation cos 2(theta)+8cos (theta)+9=0 in terms of cos (theta) and show that for cos(theta), it has equal roots. Show that there are no real roots for (theta)

Question 826309: Write the equation cos 2(theta)+8cos (theta)+9=0 in terms of cos (theta) and show that for cos(theta), it has equal roots.
Show that there are no real roots for (theta)
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

To rewrite this in terms of we will need to replace the . There are three variations of this formula:

Any of them will do but since we only want cos, not sin, we will use the one in the middle:

Combining the -1 and +9:

We can make this simpler if we divide both sides by 2 (since 2 is a factor of every term):

The equation we now have is a quadratic equation for . So we can use the discriminant, , to terminate the number and types of roots for . With a "a" of 1, a "b" of 4 and a "c" of 4 we get:

A discriminant of 0 means: two equal real roots.

To determine what the roots for are, we need to solve the equation. The quadratic equation fits the pattern with "a" equal to and "b" equal to 2. So the equation factors according to the pattern:

From the Zero Product Property:

Subtracting 2 we get:

Here we run into a problem. All cos's have values between -1 and 1. It is not possible for cos of anything to be a -2. So there is no solution for your equation.
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