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

The general solution for csc^2(x)=sin(x) is x= pi/2+2pin

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csc^2(x)=sin(x)

Solution

csc^{2} (x) = sin (x)

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

csc^{2} (x) = sin (x)

Subtract

sin (x)

from both sides

csc^{2} (x) - sin (x) = 0

Rewrite using trig identities

csc^{2} (x) - sin (x)

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

= csc^{2} (x) - \frac{1}{csc ( x )}

csc^{2} (x) - \frac{1}{csc ( x )} = 0

Solve by substitution

csc^{2} (x) - \frac{1}{csc ( x )} = 0

Let:

csc (x) = u

u^{2} - \frac{1}{u} = 0

u^{2} - \frac{1}{u} = 0 : u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{2} - \frac{1}{u} = 0

Multiply both sides by

u

u^{2} - \frac{1}{u} = 0

Multiply both sides by

u

u^{2} u - \frac{1}{u} u = 0 \cdot u

Simplify

u^{2} u - \frac{1}{u} u = 0 \cdot u

Simplify

u^{2} u : u^{3}

u^{2} u

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= u^{2 + 1}

Add the numbers:

2 + 1 = 3

= u^{3}

Simplify

- \frac{1}{u} u : - 1

- \frac{1}{u} u

Multiply fractions:

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

= - \frac{1 \cdot u}{u}

Cancel the common factor:

u

= - 1

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u^{3} - 1 = 0

u^{3} - 1 = 0

u^{3} - 1 = 0

Solve

u^{3} - 1 = 0 : u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{3} - 1 = 0

Move

1

to the right side

u^{3} - 1 = 0

Add

1

to both sides

u^{3} - 1 + 1 = 0 + 1

Simplify

u^{3} = 1

u^{3} = 1

For

x^{3} = f (a)

the solutions are

u = 1, u = \frac{- 1 + 3 i}{2}, u = \frac{- 1 - 3 i}{2}

Simplify

\frac{- 1 + 3 i}{2} : - \frac{1}{2} + i \frac{3}{2}

\frac{- 1 + 3 i}{2}

Rewrite

\frac{- 1 + 3 i}{2}

in standard complex form:

- \frac{1}{2} + \frac{3}{2} i

\frac{- 1 + 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{2} = - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3}{2} i

Simplify

\frac{- 1 - 3 i}{2} : - \frac{1}{2} - i \frac{3}{2}

\frac{- 1 - 3 i}{2}

Rewrite

\frac{- 1 - 3 i}{2}

in standard complex form:

- \frac{1}{2} - \frac{3}{2} i

\frac{- 1 - 3 i}{2}

Apply the fraction rule:

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

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

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

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

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u^{2} - \frac{1}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Substitute back

u = csc (x)

csc (x) = 1, csc (x) = - \frac{1}{2} + i \frac{3}{2}, csc (x) = - \frac{1}{2} - i \frac{3}{2}

csc (x) = 1, csc (x) = - \frac{1}{2} + i \frac{3}{2}, csc (x) = - \frac{1}{2} - i \frac{3}{2}

csc (x) = 1 : x = \frac{π}{2} + 2 πn

csc (x) = 1

General solutions for

csc (x) = 1

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

csc (x) = - \frac{1}{2} + i \frac{3}{2} :

No Solution

csc (x) = - \frac{1}{2} + i \frac{3}{2}

No Solution

csc (x) = - \frac{1}{2} - i \frac{3}{2} :

No Solution

csc (x) = - \frac{1}{2} - i \frac{3}{2}

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for csc^2(x)=sin(x) ?
The general solution for csc^2(x)=sin(x) is x= pi/2+2pin