Solve 2cos2a-5cosa=4 for 0°<=a<=360°?

To solve the equation 2cos(2a) - 5cos(a) = 4 for 0° <= a <= 360°, we can use the double angle formula for cosine:

cos(2a) = 2cos^2(a) - 1

Substitute this into the equation:

2(2cos^2(a) - 1) - 5cos(a) = 4
4cos^2(a) - 2 - 5cos(a) = 4
4cos^2(a) - 5cos(a) - 6 = 0

Now, let's solve this quadratic equation for cos(a):

4cos^2(a) - 8cos(a) + 3cos(a) - 6 = 0
4cos(a)(cos(a) - 2) + 3(cos(a) - 2) = 0
(4cos(a) + 3)(cos(a) - 2) = 0

This gives us two possible solutions:

1) 4cos(a) + 3 = 0
cos(a) = -3/4

This solution is not valid as the cosine function only takes values between -1 and 1.

2) cos(a) - 2 = 0
cos(a) = 2
This solution is also not valid as the cosine function only takes values between -1 and 1.

Therefore, the equation 2cos(2a) - 5cos(a) = 4 has no solution for 0° <= a <= 360°.