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

The general solution for 2sin(x)=sqrt(3),0<= x<= 2pi is x= pi/3 ,x=(2pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(x)=sqrt(3),0<= x<= 2pi

2 sin (x) = 3, 0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{2 π}{3}

Degrees

x = 6 0^{\circ}, x = 12 0^{\circ}

Solution steps

2 sin (x) = 3, 0 \leq x \leq 2 π

Divide both sides by

2

2 sin (x) = 3

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{2 π}{3}

x, z = arcsin (x + y)

150 cos (x - 8) = 0

8 sin^{2} (x) - 6 sin (x) + 1 = 0

sin (5 x) - cos (3 0^{\circ}) = 0

2 cos^{2} (t) + 2 cos (2 t) = 0

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