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

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

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sin(x)= 3/(2sqrt(3))

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

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

Solution steps

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

Simplify

\frac{3}{2 3} : \frac{3}{2}

\frac{3}{2 3}

Apply radical rule:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3}{2 \cdot 3 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= \frac{3 ^{1 - \frac{1}{2}}}{2}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{3 ^{\frac{1}{2}}}{2}

Apply radical rule:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \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

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

tan (x) = \frac{3}{10}

arccos (2 x) + arccos (x) = \frac{π}{3}

1.74 \cdot sin (40.3 1^{\circ}) = 1.33 \cdot sin (θ)

5 sin (θ) = 5 cos (θ)

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