What is the general solution for 1/2-cos(x)=0 ?

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

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1/2-cos(x)=0

\frac{1}{2} - cos (x) = 0

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

Degrees

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

Solution steps

\frac{1}{2} - cos (x) = 0

Move

\frac{1}{2}

to the right side

\frac{1}{2} - cos (x) = 0

Subtract

\frac{1}{2}

from both sides

\frac{1}{2} - cos (x) - \frac{1}{2} = 0 - \frac{1}{2}

Simplify

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

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

Divide both sides by

- 1

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

Divide both sides by

- 1

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

Simplify

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

Simplify

\frac{- cos ( x )}{- 1} : cos (x)

\frac{- cos ( x )}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

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

Apply rule

\frac{a}{1} = a

= cos (x)

Simplify

\frac{- \frac{1}{2}}{- 1} : \frac{1}{2}

\frac{- \frac{1}{2}}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{\frac{1}{2}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{1}{2}

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

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

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

cos (θ) sin (θ) = 1

x, 4 sin (x) + 2 = 0

3 = - tan (x)

- tan^{2} (x) - 2 tan (x) - 1 = 0

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