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

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

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1-|cos(x)|=0

Solution

1 - ∣ cos (x) ∣ = 0

Solution

x = π + 2 πn, x = 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

1 - ∣ cos (x) ∣ = 0

Solve by substitution

1 - ∣ cos (x) ∣ = 0

Let:

cos (x) = u

1 - ∣ u ∣ = 0

1 - ∣ u ∣ = 0 : u = - 1 or u = 1

1 - ∣ u ∣ = 0

Move

1

to the right side

1 - ∣ u ∣ = 0

Subtract

1

from both sides

1 - ∣ u ∣ - 1 = 0 - 1

Simplify

- ∣ u ∣ = - 1

- ∣ u ∣ = - 1

Divide both sides by

- 1

- ∣ u ∣ = - 1

Divide both sides by

- 1

\frac{- ∣ u ∣}{- 1} = \frac{- 1}{- 1}

Simplify

∣ u ∣ = 1

∣ u ∣ = 1

Apply absolute rule: If

∣ u ∣ = a, a > 0

then

u = a

u = - a

u = - 1 or u = 1

Substitute back

u = cos (x)

cos (x) = - 1 or cos (x) = 1

cos (x) = - 1 or cos (x) = 1

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = π + 2 πn, x = 2 πn

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Frequently Asked Questions (FAQ)

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