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 < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Solution steps

2 cos (θ) = 1, 0 \leq θ < 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 < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos^{2} (x) - tan (x) cos^{2} (x) = 0

3 tan (x) - 1 = 0, 0 \leq x \leq 2 π

6 + 12 sin (θ) = 0

t, cos (2 t) = sin (t)

tan (x) = 1.8

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