What is the general solution for solvefor y,(sqrt(3))/2 =sin(y) ?

The general solution for solvefor y,(sqrt(3))/2 =sin(y) is y= pi/3+2pin,y=(2pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,(sqrt(3))/2 =sin(y)

solve for

y, \frac{3}{2} = sin (y)

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

Degrees

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

Solution steps

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

Switch sides

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

General solutions for

sin (y) = \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}

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

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

tan (β) = 2, 18 0^{\circ} \leq β \leq 27 0^{\circ}

cos (x) = \frac{1}{3}, cos (2 x)

3 sin (2 x) \cdot cos (x) = 2 sin^{3} (x)

x, sin (x) \cdot cos (y) = 0

2 cos (3 x - 1) = - 1

What is the general solution for solvefor y,(sqrt(3))/2 =sin(y) ?
The general solution for solvefor y,(sqrt(3))/2 =sin(y) is y= pi/3+2pin,y=(2pi)/3+2pin