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

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

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

2 sin (3 x) = 2, 0 \leq x \leq 2 π

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

Degrees

x = 3 0^{\circ}, x = 15 0^{\circ}, x = 27 0^{\circ}

Solution steps

2 sin (3 x) = 2, 0 \leq x \leq 2 π

Divide both sides by

2

2 sin (3 x) = 2

Divide both sides by

2

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

Simplify

sin (3 x) = 1

sin (3 x) = 1

General solutions for

sin (3 x) = 1

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}

3 x = \frac{π}{2} + 2 πn

3 x = \frac{π}{2} + 2 πn

Solve

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

3 x = \frac{π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

\frac{\frac{π}{2}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

Solutions for the range

0 \leq x \leq 2 π

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

7.9615 = 59.2153 cos (x) - 12 sin (x)

1 - tan (x) = 0

2 cos^{2} (x) - 3 cos (x) = 0, 0^{\circ} < x < 18 0^{\circ}

\frac{tan ( x )}{cos ( x )} = \frac{2}{3}

tan (x) + 2 cos (x) csc (x) = sec (x) cos (x) + cot (x)

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