What is the general solution for sin(2x)=(sqrt(3))/2 ,0<= x<= 360 ?

Q: What is the general solution for sin(2x)=(sqrt(3))/2 ,0<= x<= 360 ?

The general solution for sin(2x)=(sqrt(3))/2 ,0<= x<= 360 is x=30,x=60,x=210,x=240

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2x)=(sqrt(3))/2 ,0<= x<= 360

Solution

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 3 0^{\circ}, x = 6 0^{\circ}, x = 21 0^{\circ}, x = 24 0^{\circ}

Radians

x = \frac{π}{6}, x = \frac{π}{3}, x = \frac{7 π}{6}, x = \frac{4 π}{3}

Solution steps

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

General solutions for

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

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 6 0^{\circ} + 36 0^{\circ} n, 2 x = 12 0^{\circ} + 36 0^{\circ} n

2 x = 6 0^{\circ} + 36 0^{\circ} n, 2 x = 12 0^{\circ} + 36 0^{\circ} n

Solve

2 x = 6 0^{\circ} + 36 0^{\circ} n : x = 3 0^{\circ} + 18 0^{\circ} n

2 x = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 x = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 3 0^{\circ} + 18 0^{\circ} n

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

\frac{6 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 3 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 3 0^{\circ} + 18 0^{\circ} n

x = 3 0^{\circ} + 18 0^{\circ} n

x = 3 0^{\circ} + 18 0^{\circ} n

x = 3 0^{\circ} + 18 0^{\circ} n

Solve

2 x = 12 0^{\circ} + 36 0^{\circ} n : x = 6 0^{\circ} + 18 0^{\circ} n

2 x = 12 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 x = 12 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 6 0^{\circ} + 18 0^{\circ} n

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{12 0 ^{\circ}}{2} = 6 0^{\circ}

\frac{12 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{36 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 6 0^{\circ}

Cancel the common factor:

2

= 6 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 6 0^{\circ} + 18 0^{\circ} n

x = 6 0^{\circ} + 18 0^{\circ} n

x = 6 0^{\circ} + 18 0^{\circ} n

x = 6 0^{\circ} + 18 0^{\circ} n

x = 3 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 3 0^{\circ}, x = 6 0^{\circ}, x = 21 0^{\circ}, x = 24 0^{\circ}

Popular Examples

3sin^2(x)+8cos^2(x)=6

3 sin^{2} (x) + 8 cos^{2} (x) = 6

tan(θ)= 2/8

tan (θ) = \frac{2}{8}

4tan^2(x)+7tan(x)-4=0,cot(2x)

4 tan^{2} (x) + 7 tan (x) - 4 = 0, cot (2 x)

tan^3(x)=tan^2(x),0<= x<= 2pi

tan^{3} (x) = tan^{2} (x), 0 \leq x \leq 2 π

11.75=5sin((pit)/6-(7pi)/6)+8

11.75 = 5 sin (\frac{π t}{6} - \frac{7 π}{6}) + 8

Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)=(sqrt(3))/2 ,0<= x<= 360 ?
The general solution for sin(2x)=(sqrt(3))/2 ,0<= x<= 360 is x=30,x=60,x=210,x=240