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)

Algebra ->  Trigonometry-basics -> 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)      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
cos%282theta%29%2B8cos%28theta%29%2B9=0
To rewrite this in terms of cos%28theta%29 we will need to replace the cos%282theta%29. There are three variations of this formula:
  • cos%282theta%29+=+cos%5E2%28theta%29-sin%5E2%28theta%29
  • cos%282theta%29+=+2cos%5E2%28theta%29-1%29
  • cos%282theta%29+=+1-2sin%5E2%28theta%29
Any of them will do but since we only want cos, not sin, we will use the one in the middle:
%282cos%5E2%28theta%29-1%29%2B8cos%28theta%29%2B9=0
Combining the -1 and +9:
2cos%5E2%28theta%29%2B8cos%28theta%29%2B8=0
We can make this simpler if we divide both sides by 2 (since 2 is a factor of every term):
cos%5E2%28theta%29%2B4cos%28theta%29%2B4=0

The equation we now have is a quadratic equation for cos%28theta%29. So we can use the discriminant, b%5E2-4ac, to terminate the number and types of roots for cos%28theta%29. With a "a" of 1, a "b" of 4 and a "c" of 4 we get:
b%5E2-4ac+=+%284%29%5E2-4%281%29%284%29+=+16+-+4%281%29%284%29+=+16+-+16+=+0
A discriminant of 0 means: two equal real roots.

To determine what the roots for theta are, we need to solve the equation. The quadratic equation fits the a%5E2%2B2ab%2Bb%5E2+=+%28a%2Bb%29%5E2 pattern with "a" equal to cos%28theta%29 and "b" equal to 2. So the equation factors according to the pattern:
%28cos%28theta%29%2B2%29%5E2+=+0
From the Zero Product Property:
cos%28theta%29%2B2+=+0
Subtracting 2 we get:
cos%28theta%29+=+-2
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.