What is the general solution for sin(3x)=1,0<= x<2pi ?

The general solution for sin(3x)=1,0<= x<2pi is x= pi/6 ,x=(5pi)/6 ,x=(3pi)/2

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sin(3x)=1,0<= x<2pi

sin (3 x) = 1, 0 \leq x < 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

Solution steps

sin (3 x) = 1, 0 \leq x < 2 π

General solutions for

sin (3 x) = 1

3 x = \frac{π}{2} + 2 πn

Solve

3 x = \frac{π}{2} + 2 πn : x = \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

\frac{2}{π} = sin (x)

1 - 3 cos (x) = 0

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

arccos (x) = arcsin (\frac{3}{5})

x, y = 2 tan (x)

What is the general solution for sin(3x)=1,0<= x<2pi ?
The general solution for sin(3x)=1,0<= x<2pi is x= pi/6 ,x=(5pi)/6 ,x=(3pi)/2