What is the general solution for sin^2(3x)=1 ?

The general solution for sin^2(3x)=1 is x= pi/6+(2pin)/3 ,x= pi/2+(2pin)/3

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sin^2(3x)=1

Solution

sin^{2} (3 x) = 1

Solution

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

Degrees

x = 3 0^{\circ} + 12 0^{\circ} n, x = 9 0^{\circ} + 12 0^{\circ} n

Solution steps

sin^{2} (3 x) = 1

Solve by substitution

sin^{2} (3 x) = 1

Let:

sin (3 x) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = sin (3 x)

sin (3 x) = 1, sin (3 x) = - 1

sin (3 x) = 1, sin (3 x) = - 1

sin (3 x) = 1 : x = \frac{π}{6} + \frac{2 πn}{3}

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}

sin (3 x) = - 1 : x = \frac{π}{2} + \frac{2 πn}{3}

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{3 π}{2} + 2 πn

3 x = \frac{3 π}{2} + 2 πn

Solve

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

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

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

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(3x)=1 ?
The general solution for sin^2(3x)=1 is x= pi/6+(2pin)/3 ,x= pi/2+(2pin)/3