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

The general solution for |sin(x)|=1 is x=(3pi)/2+2pin,x= pi/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

|sin(x)|=1

∣ sin (x) ∣ = 1

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

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

∣ sin (x) ∣ = 1

Solve by substitution

∣ sin (x) ∣ = 1

Let:

sin (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 = sin (x)

sin (x) = - 1 or sin (x) = 1

sin (x) = - 1 or sin (x) = 1

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

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

tan (x) csc (x) = 2

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

sin^{2} (x) - 5 sin (x) + 6 = 0

4 = 3 - cot (x) - csc (x)

7 sin (x) tan (x) = 3 tan (x)

What is the general solution for |sin(x)|=1 ?
The general solution for |sin(x)|=1 is x=(3pi)/2+2pin,x= pi/2+2pin