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

The general solution for 1=sin(x^2) is x=sqrt((pi+4pin)/2),x=-sqrt((pi+4pin)/2)

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1=sin(x^2)

1 = sin (x^{2})

x = \frac{π + 4 πn}{2}, x = - \frac{π + 4 πn}{2}

Degrees

x = 0^{\circ} + 160.57117 \dots^{\circ} n, x = 0^{\circ} - 160.57117 \dots^{\circ} n

Solution steps

1 = sin (x^{2})

Switch sides

sin (x^{2}) = 1

General solutions for

sin (x^{2}) = 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}

x^{2} = \frac{π}{2} + 2 πn

x^{2} = \frac{π}{2} + 2 πn

Solve

x^{2} = \frac{π}{2} + 2 πn : x = \frac{π + 4 πn}{2}, x = - \frac{π + 4 πn}{2}

x^{2} = \frac{π}{2} + 2 πn

For

x^{2} = f (a)

the solutions are

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

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

Simplify

\frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

\frac{π}{2} + 2 πn

Join

\frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

\frac{π}{2} + 2 πn

Convert element to fraction:

2 πn = \frac{2 πn 2}{2}

= \frac{π}{2} + \frac{2 πn \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 2 πn \cdot 2}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π + 4 πn}{2}

= \frac{π + 4 πn}{2}

Simplify

- \frac{π}{2} + 2 πn : - \frac{π + 4 πn}{2}

- \frac{π}{2} + 2 πn

Join

\frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

\frac{π}{2} + 2 πn

Convert element to fraction:

2 πn = \frac{2 πn 2}{2}

= \frac{π}{2} + \frac{2 πn \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 2 πn \cdot 2}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π + 4 πn}{2}

= - \frac{π + 4 πn}{2}

x = \frac{π + 4 πn}{2}, x = - \frac{π + 4 πn}{2}

x = \frac{π + 4 πn}{2}, x = - \frac{π + 4 πn}{2}

What is the general solution for 1=sin(x^2) ?
The general solution for 1=sin(x^2) is x=sqrt((pi+4pin)/2),x=-sqrt((pi+4pin)/2)