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

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

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2cos(x)=1,0<= x<2pi

2 cos (x) = 1, 0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Degrees

x = 6 0^{\circ}, x = 30 0^{\circ}

Solution steps

2 cos (x) = 1, 0 \leq x < 2 π

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{1}{2}

Simplify

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = - \frac{3}{2}

tan^{2} (x) + 2 tan (x) - 3 = 0

sin (2 x - \frac{π}{3}) = - \frac{1}{2}

2 cos (α) = 3 tan (α)

3 tan^{2} (θ) + 7 tan (θ) - 2 = - 4

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