Question

help please finding exact solutions of cos2x+10cosx=5

Answer 1

Is this the problem?
$$
\cos(2x)+10\cos x=5
$$

Answer 2

yes and you have to find solutions

Answer 3

i tryed to use the identity for cos(2x) but it didnt work

Answer 4

There are real and imaginary solutions:
http://www.wolframalpha.com/input/?i=cos%282x%29%2B10cosx%3D5

Here are the details for the real solution:
Solve for x:
10 cos(x)+cos(2 x) = 5
Simplify the left hand side.
Transform 10 cos(x)+cos(2 x) into a polynomial with respect to cos(x) using cos(2 x) = 2 cos^2(x)-1:
-1+10 cos(x)+2 cos^2(x) = 5
Write the quadratic equation in standard form.
Divide both sides by 2:
-1/2+5 cos(x)+cos^2(x) = 5/2
Solve the quadratic equation by completing the square.
Add 1/2 to both sides:
5 cos(x)+cos^2(x) = 3
Take one half of the coefficient of cos(x) and square it, then add it to both sides.
Add 25/4 to both sides:
25/4+5 cos(x)+cos^2(x) = 37/4
Factor the left hand side.
Write the left hand side as a square:
(5/2+cos(x))^2 = 37/4
Eliminate the exponent on the left hand side.
Take the square root of both sides:
5/2+cos(x) = sqrt(37)/2 or 5/2+cos(x) = -sqrt(37)/2
Look at the first equation: Solve for cos(x).
Subtract 5/2 from both sides:
cos(x) = sqrt(37)/2-5/2 or 5/2+cos(x) = -sqrt(37)/2
Eliminate the cosine from the left hand side.
Take the inverse cosine of both sides:
x = cos^(-1)(sqrt(37)/2-5/2)+2 pi n_1 for n_1 element Z or x = 2 pi n_2-cos^(-1)(sqrt(37)/2-5/2) for n_2 element Z
or 5/2+cos(x) = -sqrt(37)/2
Look at the third equation: Solve for cos(x).
Subtract 5/2 from both sides:
x = cos^(-1)(sqrt(37)/2-5/2)+2 pi n_1 for n_1 element Z
or x = 2 pi n_2-cos^(-1)(sqrt(37)/2-5/2) for n_2 element Z
or cos(x) = -5/2-sqrt(37)/2
Show the equation has no solution.
cos(x) = -5/2-sqrt(37)/2 has no solution since for all x element C, -1<=Re(cos(x))<=1 and -5/2-sqrt(37)/2<-1:
Answer: |
| x = cos^(-1)(sqrt(37)/2-5/2)+2 pi n_1 for n_1 element Z
or x = 2 pi n_2-cos^(-1)(sqrt(37)/2-5/2) for n_2 element Z