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

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

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|cos(2x)|=1

∣ cos (2 x) ∣ = 1

x = \frac{π}{2} + πn, x = πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 18 0^{\circ} n

Solution steps

∣ cos (2 x) ∣ = 1

Solve by substitution

∣ cos (2 x) ∣ = 1

Let:

cos (2 x) = u

∣ u ∣ = 1

∣ u ∣ = 1 : u = - 1 or 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 (2 x)

cos (2 x) = - 1 or cos (2 x) = 1

cos (2 x) = - 1 or cos (2 x) = 1

cos (2 x) = - 1 : x = \frac{π}{2} + πn

cos (2 x) = - 1

General solutions for

cos (2 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}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

cos (2 x) = 1 : x = πn

cos (2 x) = 1

General solutions for

cos (2 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}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

Combine all the solutions

x = \frac{π}{2} + πn, x = πn

sin^{3} (x) = sin^{3} (x) tan (x)

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

2 cos^{2} (4 x) = 1

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

csc^{2} (x) + 3 csc (x) + 2 = 0

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