What is the general solution for 2cos^2(θ)-cos(θ)=1,0<= θ<= 2pi ?

The general solution for 2cos^2(θ)-cos(θ)=1,0<= θ<= 2pi is θ=0,θ=2pi,θ=(2pi)/3 ,θ=(4pi)/3

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2cos^2(θ)-cos(θ)=1,0<= θ<= 2pi

2 cos^{2} (θ) - cos (θ) = 1, 0 \leq θ \leq 2 π

θ = 0, θ = 2 π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Solution steps

2 cos^{2} (θ) - cos (θ) = 1, 0 \leq θ \leq 2 π

Solve by substitution

cos (θ) = 1, cos (θ) = - \frac{1}{2}

cos (θ) = 1, 0 \leq θ \leq 2 π : θ = 0, θ = 2 π

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

Combine all the solutions

θ = 0, θ = 2 π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

tan (x) - 3 = 0, 0 \leq x \leq 2 π

1 = tan (0 + c)

4 sin^{2} (x) = 3 - 4 sin (x)

tan (θ) = - \frac{2}{2}

\frac{sin ( 7 8 ^{\circ} )}{14} = \frac{sin ( a )}{6}

What is the general solution for 2cos^2(θ)-cos(θ)=1,0<= θ<= 2pi ?
The general solution for 2cos^2(θ)-cos(θ)=1,0<= θ<= 2pi is θ=0,θ=2pi,θ=(2pi)/3 ,θ=(4pi)/3