What is the general solution for 2cos(t)+1=0 ?

The general solution for 2cos(t)+1=0 is t=(2pi)/3+2pin,t=(4pi)/3+2pin

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2cos(t)+1=0

2 cos (t) + 1 = 0

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Degrees

t = 12 0^{\circ} + 36 0^{\circ} n, t = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (t) + 1 = 0

Move

1

to the right side

2 cos (t) + 1 = 0

Subtract

1

from both sides

2 cos (t) + 1 - 1 = 0 - 1

Simplify

2 cos (t) = - 1

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{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

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

tan^{2} (θ) = - tan (θ)

sec (θ) = - \frac{2}{3}

tan (θ) = \frac{8}{12}

sin (t) + cos (t) = 0

What is the general solution for 2cos(t)+1=0 ?
The general solution for 2cos(t)+1=0 is t=(2pi)/3+2pin,t=(4pi)/3+2pin