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

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

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2cos(θ)=1,0<= θ<= 2pi

2 cos (θ) = 1, 0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Solution steps

2 cos (θ) = 1, 0 \leq θ \leq 2 π

Divide both sides by

2

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

arctan (x + 1) + arctan (x - 1) = arctan (\frac{8}{31})

4 cos (2 θ) + 19 = - 22 cos (θ) + 6

8 cos^{2} (x) + 16 cos (x) + 8 = 0

cos (x) = - 0.35

cos (x) = - 0.25

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