What is the general solution for sin(3x)=-1,0<= x<= 2pi ?

The general solution for sin(3x)=-1,0<= x<= 2pi is x= pi/2 ,x=(7pi)/6 ,x=(11pi)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(3x)=-1,0<= x<= 2pi

sin (3 x) = - 1, 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

Degrees

x = 9 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}

Solution steps

sin (3 x) = - 1, 0 \leq x \leq 2 π

General solutions for

sin (3 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}

3 x = \frac{3 π}{2} + 2 πn

3 x = \frac{3 π}{2} + 2 πn

Solve

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

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{3} + \frac{2 πn}{3} : \frac{π}{2} + \frac{2 πn}{3}

\frac{\frac{3 π}{2}}{3} + \frac{2 πn}{3}

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (θ) = 0.25

3 0^{2} = 2 0^{2} + 1 8^{2} - 2 (20) (18) cos (A)

cos (x) = \frac{308}{1475}

tan (θ) = \frac{6}{3}

csc (θ) = - \frac{13}{5}

What is the general solution for sin(3x)=-1,0<= x<= 2pi ?
The general solution for sin(3x)=-1,0<= x<= 2pi is x= pi/2 ,x=(7pi)/6 ,x=(11pi)/6