Solve Cos^2(x)+cos(x)=cos(2x). Give exact answers within the interval [0,2π)
Ive got the equation down to -cos^2(x)+cos(x)+1=0 or and it can be simplified too sin^2(x)+cos(x)=0
If you could tell me where to go from either of these two, it would be great help.. Thankyou for any responses
cos^2(x) + cos(x) = 2cos^2(x) - 1
cos^2(x) - cos(x) - 1 = 0
This is just like
u^2 - u - 1 = 0
u = [1 +/- sqrt(5)]/2
since u = cos(x) and must be less than 1,
cos(x) = (1-sqrt(5))/2
will be in QII or QIII
To solve the equation cos^2(x) + cos(x) = cos(2x) within the interval [0, 2π), you have simplified it correctly to sin^2(x) + cos(x) = 0.
One way to progress is by using the identity sin^2(x) + cos^2(x) = 1. By subtracting cos^2(x) from both sides, we can rewrite the equation as:
sin^2(x) + cos(x) - cos^2(x) = 1 - cos^2(x)
Now, we can substitute 1 - cos^2(x) with sin^2(x) by using the identity mentioned above. The equation becomes:
sin^2(x) + cos(x) - cos^2(x) = sin^2(x)
Rearranging the terms yields:
sin^2(x) - cos^2(x) + cos(x) = 0
Now, we can factor the left side of the equation:
(sin(x) - cos(x))(sin(x) + cos(x)) + cos(x) = 0
Now we have two possibilities:
1. (sin(x) - cos(x)) = 0
Solving this equation, we get sin(x) = cos(x). In the interval [0, 2π), the solutions are x = π/4 and x = 5π/4.
2. (sin(x) + cos(x)) + cos(x) = 0
Expanding, we get sin(x) + 2cos(x) = 0. Solving this equation, we find that x = (3π/2) or x = (7π/2), which are not within the given interval [0, 2π).
Therefore, the solutions within the interval [0, 2π) are x = π/4 and x = 5π/4.