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

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

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2cos(t)=1

2 cos (t) = 1

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

Degrees

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

Solution steps

2 cos (t) = 1

Divide both sides by

2

2 cos (t) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (t) = \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}

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

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

cos (x) = \frac{2}{13}

cos (θ) = sec (θ)

4.45 tan^{2} (θ) - 23 tan (θ) + 16.15 = 0

sin^{2} (x) = \frac{( 3 m - 2 )}{8}

arccos (x) = arcsin (\frac{1}{3}) + arccos (\frac{1}{4})

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