What is the general solution for 2cos(a)=1 ?

The general solution for 2cos(a)=1 is a= pi/3+2pin,a=(5pi)/3+2pin

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2cos(a)=1

2 cos (a) = 1

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

Degrees

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (a) = 1

Divide both sides by

2

2 cos (a) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (a) = \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}

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

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

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

3 sin (\frac{π}{2} x) = 2

3 cos (A) - 1 = 0

cot (\frac{x}{2}) = - 1

5 sin (\frac{2 x - π}{2}) = 0.5

What is the general solution for 2cos(a)=1 ?
The general solution for 2cos(a)=1 is a= pi/3+2pin,a=(5pi)/3+2pin