What is the general solution for 2sin(3x)+sqrt(2)=0 ?

The general solution for 2sin(3x)+sqrt(2)=0 is x=(5pi}{12}+(2pin)/3 ,x=(7pi)/(12)+\frac{2pin)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(3x)+sqrt(2)=0

Solution

2 sin (3 x) + 2 = 0

Solution

x = \frac{5 π}{12} + \frac{2 πn}{3}, x = \frac{7 π}{12} + \frac{2 πn}{3}

Degrees

x = 7 5^{\circ} + 12 0^{\circ} n, x = 10 5^{\circ} + 12 0^{\circ} n

Solution steps

2 sin (3 x) + 2 = 0

Move

2

to the right side

2 sin (3 x) + 2 = 0

Subtract

2

from both sides

2 sin (3 x) + 2 - 2 = 0 - 2

Simplify

2 sin (3 x) = - 2

2 sin (3 x) = - 2

Divide both sides by

2

2 sin (3 x) = - 2

Divide both sides by

2

\frac{2 sin ( 3 x )}{2} = \frac{- 2}{2}

Simplify

sin (3 x) = - \frac{2}{2}

sin (3 x) = - \frac{2}{2}

General solutions for

sin (3 x) = - \frac{2}{2}

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{5 π}{4} + 2 πn, 3 x = \frac{7 π}{4} + 2 πn

3 x = \frac{5 π}{4} + 2 πn, 3 x = \frac{7 π}{4} + 2 πn

Solve

3 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{12} + \frac{2 πn}{3}

3 x = \frac{5 π}{4} + 2 πn

Divide both sides by

3

3 x = \frac{5 π}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{5 π}{4}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{5 π}{4}}{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{5 π}{4}}{3} + \frac{2 πn}{3} : \frac{5 π}{12} + \frac{2 πn}{3}

\frac{\frac{5 π}{4}}{3} + \frac{2 πn}{3}

\frac{\frac{5 π}{4}}{3} = \frac{5 π}{12}

\frac{\frac{5 π}{4}}{3}

Apply the fraction rule:

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

= \frac{5 π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{5 π}{12}

= \frac{5 π}{12} + \frac{2 πn}{3}

x = \frac{5 π}{12} + \frac{2 πn}{3}

x = \frac{5 π}{12} + \frac{2 πn}{3}

x = \frac{5 π}{12} + \frac{2 πn}{3}

Solve

3 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{12} + \frac{2 πn}{3}

3 x = \frac{7 π}{4} + 2 πn

Divide both sides by

3

3 x = \frac{7 π}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{7 π}{4}}{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{7 π}{4}}{3} + \frac{2 πn}{3} : \frac{7 π}{12} + \frac{2 πn}{3}

\frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

\frac{\frac{7 π}{4}}{3} = \frac{7 π}{12}

\frac{\frac{7 π}{4}}{3}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{7 π}{12}

= \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{5 π}{12} + \frac{2 πn}{3}, x = \frac{7 π}{12} + \frac{2 πn}{3}

Popular Examples

0=cos(θ)

0 = cos (θ)

sec^2(x)+3tan(x)=5

sec^{2} (x) + 3 tan (x) = 5

2sin(θ)+1=csc(θ)

2 sin (θ) + 1 = csc (θ)

cot(x)-1=csc(x)

cot (x) - 1 = csc (x)

-6cos^2(θ)-3cos(θ)+1=-4cos(θ)

- 6 cos^{2} (θ) - 3 cos (θ) + 1 = - 4 cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(3x)+sqrt(2)=0 ?
The general solution for 2sin(3x)+sqrt(2)=0 is x=(5pi}{12}+(2pin)/3 ,x=(7pi)/(12)+\frac{2pin)/3