Expert: Ahmed Salami - 2/15/2010

Question
Hello Ahmed

Can you please help in solving 2 cos^2 (theta) + cos (theta) + 1 = 0; -360degrees </= theta </= 360degrees.

Thank you

Answer
Hi Freddie,
2cos²θ + cosθ + 1 = 0
is a quadratic equation in cosθ which is easily solved by setting x = cosθ. The equation becomes
2x² + x + 1 = 0
and this happens to have complex (i.e no real) solutions.

But just as an illustration of the method of solution, lets say the equation was
2cos²θ + cosθ - 1 = 0
It becomes
2x² + x - 1 = 0
2x² + 2x - x - 1 = 0
2x(x+1) - (x+1) = 0
(x+1)(2x-1) = 0
x = -1 or 1/2
cosθ = -1 or 1/2
For cosθ = -1 and -360° ≤ θ ≤ 360°
θ = -180° or 180°
For cosθ = 1/2 and -360° ≤ θ ≤ 360°
θ = -300°, -60°, 60° or 300°

Regards