How do you solve $2 {cos}^{2} (x) + cos (x) - 1 = 0$ $2 cos^2 (x) + cos(x) - 1 = 0$ from [0,2pi]?

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Answer 1 · 2015-09-21T09:13:47

x= $\frac{π}{3}, \frac{5 π}{3}$ $pi/3, (5pi)/3$

Solving the quadratic expression in cos x, it would be $cos x = \frac{- 1 \pm \sqrt{1 + 8}}{2}$ $cosx = (-1+- sqrt(1+8))/2$

$cos x = \frac{- 1 \pm 3}{2}$ $cos x = (-1+-3)/2$

$cos x = 1, - 2$ $cos x= 1, -2$ cos x is not feasible, hence it is cos x= $\frac{1}{2}$ $1/2$ , x= $\frac{π}{3}, \frac{5 π}{3}$ $pi/3, (5pi)/3$

How do you solve 2 cos^2 (x) + cos(x) - 1 = 0 2cos2(x)+cos(x)−1=0 2 cos^2 (x) + cos(x) - 1 = 0 from [0,2pi]?