What is the general solution for 2cos^2(θ)+cos(θ)=0,0<= θ<2pi ?

The general solution for 2cos^2(θ)+cos(θ)=0,0<= θ<2pi is θ= pi/2 ,θ=(3pi)/2 ,θ=(2pi)/3 ,θ=(4pi)/3

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2cos^2(θ)+cos(θ)=0,0<= θ<2pi

2 cos^{2} (θ) + cos (θ) = 0, 0 \leq θ < 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Solution steps

2 cos^{2} (θ) + cos (θ) = 0, 0 \leq θ < 2 π

Solve by substitution

cos (θ) = 0, cos (θ) = - \frac{1}{2}

cos (θ) = 0, 0 \leq θ < 2 π : θ = \frac{π}{2}, θ = \frac{3 π}{2}

cos (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

3 cos (2 x) + sin (2 x) = 0

cos (\frac{- π}{2} + \frac{x}{2}) = 0

cos (x) = - 0.175

sec^{2} (x) = 2 tan^{2} (x)

\frac{sin ( x )}{cos ( x )} - 5 + \frac{4 cos ( x )}{sin ( x )} = 0

What is the general solution for 2cos^2(θ)+cos(θ)=0,0<= θ<2pi ?
The general solution for 2cos^2(θ)+cos(θ)=0,0<= θ<2pi is θ= pi/2 ,θ=(3pi)/2 ,θ=(2pi)/3 ,θ=(4pi)/3