What are the exact roots for $2 cos θ - \sqrt{3} = 0$ $2costheta-sqrt3=0$ in $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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Answer 1 · 2018-07-27T06:00:56

The solutions are $S = {\frac{π}{6}, \frac{11}{6} π}$ $S={pi/6, 11/6pi}$

The equation is

$2 cos θ - \sqrt{3} = 0$ $2costheta-sqrt3=0$

$\Rightarrow$ $=>$ , $cos θ = \frac{\sqrt{3}}{2}$ $costheta=sqrt3/2$

The solutions are

$\Rightarrow$ $=>$ , ${(theta=pi/6),(theta=11/6pi):}$

Answer 2 · 2018-07-27T06:08:59

$color(brown)(theta = (pi/6, 11pi/6), 0 <= theta <= 2pi$

$2 cos theta - sqrt3 = 0$

$2 cos theta = sqrt3$

$cos theta = sqrt3 / 2$

![ www.s2004268.students.dlszobel.edu.ph )

$color(indigo)(theta = cos^-1 (sqrt3 / 2) = 30^@, 330^@, color(crimson)(" as cosine is positive in I and IV Quadrants."$

What are the exact roots for 2costheta-sqrt3=02cosθ−√3=02costheta-sqrt3=0 in 0<=theta<2pi0≤θ<2π0<=theta<2pi?