How do you solve the equation $2 {cos}^{2} θ - 3 cos θ + 1 = 0$ $2cos^2theta-3costheta+1=0$ for $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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Answer 1 · 2016-11-21T11:29:07

The solutions are $S = {0, \frac{π}{3}, \frac{5 π}{3}}$ $S={0,pi/3,(5pi)/3}$

Let's factorise the LHS

$2 {cos}^{2} θ - 3 cos θ + 1 = (2 cos θ - 1) (cos θ - 1) = 0$ $2cos^2theta-3costheta+1=(2costheta-1)(costheta-1)=0$

Therefore, $2 cos θ - 1 = 0$ $2costheta-1=0$

$cos θ = \frac{1}{2}$ $costheta=1/2$

and $cos θ - 1 = 0$ $costheta-1=0$

$cos θ = 1$ $cos theta=1$

We are looking for $θ \in [0, 2 π [$ $theta in [0, 2pi[$

When $cos θ = \frac{1}{2}$ $costheta=1/2$

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

When $cos θ = 1$ $costheta=1$

$θ = 0$ $theta=0$

How do you solve the equation 2cos^2theta-3costheta+1=02cos2θ−3cosθ+1=02cos^2theta-3costheta+1=0 for 0<=theta<2pi0≤θ<2π0<=theta<2pi?