solve the equation:
4cos2*theta + 3cos*theta + 3 = 0 for values between 0 and 180 degrees
If that is cosine squared in the first term, you have a quadratic equation.
Normally cosine squared would be written in ASCII as cos^2 Theta.
To solve the equation 4cos^2(theta) + 3cos(theta) + 3 = 0 for values between 0 and 180 degrees, we can use a substitution to simplify the equation. Let's denote cos(theta) as x.
So, the equation becomes:
4x^2 + 3x + 3 = 0
To solve this quadratic equation, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 4, b = 3, and c = 3. Substituting these values into the quadratic formula, we get:
x = (-(3) ± √(3^2 - 4(4)(3))) / (2(4))
= (-3 ± √(9 - 48)) / 8
= (-3 ± √(-39)) / 8
Observing that the discriminant, b^2 - 4ac, is negative (-39), we can conclude that the equation has no real solutions in the given domain of 0 to 180 degrees, because the cosine function only takes real values between -1 and 1.
Therefore, in the given interval, there are no values of theta that satisfy the equation 4cos^2(theta) + 3cos(theta) + 3 = 0.