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

The general solution for cos(2θ)=-1/2 ,0<= θ<= 2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3

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cos(2θ)=-1/2 ,0<= θ<= 2pi

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

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Solution steps

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

General solutions for

cos (2 θ) = - \frac{1}{2}

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

Solve

2 θ = \frac{2 π}{3} + 2 πn : θ = \frac{π}{3} + πn

Solve

2 θ = \frac{4 π}{3} + 2 πn : θ = \frac{2 π}{3} + πn

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

sin (2 x) = \frac{2}{5}

sin (2 x) + cos (x) = 0, 0 \leq x \leq 2 π

cos (θ) = - 0.5

sec (x) = \frac{5}{3}

x, y = \frac{sec ( x )}{cos ^{1} ( x )} + 8 y^{2}

What is the general solution for cos(2θ)=-1/2 ,0<= θ<= 2pi ?
The general solution for cos(2θ)=-1/2 ,0<= θ<= 2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3