Asked by Raul
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.
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.
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