Question #b4d8f

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Question #b4d8f

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Answer 1 · 2017-05-31T10:33:41

$\frac{2 π}{3}, \frac{4 π}{3}$ $(2pi)/3,(4pi)/3$

Explanation:

$2 cos θ + 1 = 0$ $2costheta+1=0$

$\Rightarrow cos θ = - \frac{1}{2}$ $rArrcostheta=-1/2$

$since the ratio is negative θ will be in the$ $"since the ratio is negative " theta" will be in the"$
$second and third quadrants$ $"second and third quadrants"$

$\Rightarrow θ = {cos}^{- 1} (\frac{1}{2}) = \frac{π}{3} \leftarrow related acute angle$ $rArrtheta=cos^-1(1/2)=pi/3larrcolor(red)" related acute angle"$

$\Rightarrow θ = (π - \frac{π}{3}) = \frac{2 π}{3} \leftarrow second quadrant$ $rArrtheta=(pi-pi/3)=(2pi)/3larrcolor(blue)" second quadrant"$

$\Rightarrow θ = (π + \frac{π}{3}) = \frac{4 π}{3} \leftarrow third quadrant$ $rArrtheta=(pi+pi/3)=(4pi)/3larrcolor(blue)" third quadrant"$

$\Rightarrow θ = \frac{2 π}{3}, \frac{4 π}{3} \to [0, 2 π)$ $rArrtheta=(2pi)/3,(4pi)/3to[0,2pi)$

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