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Popular Trigonometry >

solvefor x,sqrt(1+sin^3(xy^2))=y

Solution

solve for

x, 1 + sin^{3} (x y^{2}) = y

Solution

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}, x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}

Radians

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 π}{y ^{2}} n, x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 π}{y ^{2}} n

Solution steps

1 + sin^{3} (x y^{2}) = y

Square both sides:

1 + sin^{3} (x y^{2}) = y^{2}

1 + sin^{3} (x y^{2}) = y

(1 + sin^{3} (x y^{2}))^{2} = y^{2}

Expand

(1 + sin^{3} (x y^{2}))^{2} : 1 + sin^{3} (x y^{2})

(1 + sin^{3} (x y^{2}))^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((1 + sin^{3} (x y^{2}))^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (1 + sin^{3} (x y^{2}))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 1 + sin^{3} (x y^{2})

1 + sin^{3} (x y^{2}) = y^{2}

1 + sin^{3} (x y^{2}) = y^{2}

Solve

1 + sin^{3} (x y^{2}) = y^{2} : sin (x y^{2}) = 3 y^{2} - 1

1 + sin^{3} (x y^{2}) = y^{2}

Move

1

to the right side

1 + sin^{3} (x y^{2}) = y^{2}

Subtract

1

from both sides

1 + sin^{3} (x y^{2}) - 1 = y^{2} - 1

Simplify

sin^{3} (x y^{2}) = y^{2} - 1

sin^{3} (x y^{2}) = y^{2} - 1

For

x^{n} = f (a)

, n is odd, the solution is

x = n f (a)

sin (x y^{2}) = 3 y^{2} - 1

sin (x y^{2}) = 3 y^{2} - 1

Verify Solutions:

sin (x y^{2}) = 3 y^{2} - 1 {y \geq 0}

Check the solutions by plugging them into

1 + sin^{3} (x y^{2}) = y

Remove the ones that don't agree with the equation.

Plug

sin (x y^{2}) = 3 y^{2} - 1 : 1 + (3 y^{2} - 1)^{3} = y \Rightarrow y \geq 0

1 + (3 y^{2} - 1)^{3} = y

Square both sides:

y^{2} = y^{2}

1 + (3 y^{2} - 1)^{3} = y

(1 + (3 y^{2} - 1)^{3})^{2} = y^{2}

Expand

(1 + (3 y^{2} - 1)^{3})^{2} : y^{2}

(1 + (3 y^{2} - 1)^{3})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((1 + (3 y^{2} - 1)^{3})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (1 + (3 y^{2} - 1)^{3})^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 1 + (3 y^{2} - 1)^{3}

Expand

1 + (3 y^{2} - 1)^{3} : y^{2}

1 + (3 y^{2} - 1)^{3}

(3 y^{2} - 1)^{3} = y^{2} - 1

(3 y^{2} - 1)^{3}

Apply radical rule:

n a = a^{\frac{1}{n}}

= ((y^{2} - 1)^{\frac{1}{3}})^{3}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (y^{2} - 1)^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

Multiply fractions:

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

= \frac{1 \cdot 3}{3}

Cancel the common factor:

3

= 1

= y^{2} - 1

= 1 + y^{2} - 1

Group like terms

= y^{2} + 1 - 1

1 - 1 = 0

= y^{2}

= y^{2}

y^{2} = y^{2}

y^{2} = y^{2}

Both sides are equal

True for all y

Verify Solutions:

y < 0

False

, y = 0

True

, y > 0

True

1 + (3 y^{2} - 1)^{3} = y

Combine domain interval with solution interval:

True for all y

Find the function intervals:

y < 0, y = 0, y > 0

1 + (3 y^{2} - 1)^{3} = y

Find the even roots arguments zeroes:

Solve

1 + 3 y^{2} - 1^{3} = 0 : y = 0

1 + (3 y^{2} - 1)^{3} = 0

Factor

1 + (3 y^{2} - 1)^{3} : (3 y^{2} - 1 + 1) ((y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1)

1 + (3 y^{2} - 1)^{3}

Rewrite

1

1^{3}

= (3 y^{2} - 1)^{3} + 1^{3}

Apply Sum of Cubes Formula:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(3 y^{2} - 1)^{3} + 1^{3} = (3 y^{2} - 1 + 1) ((3 y^{2} - 1)^{2} - 3 y^{2} - 1 + 1)

= (3 y^{2} - 1 + 1) ((3 y^{2} - 1)^{2} - 3 y^{2} - 1 + 1)

(3 y^{2} - 1)^{2} = (y^{2} - 1)^{\frac{2}{3}}

(3 y^{2} - 1)^{2}

Apply radical rule:

n a = a^{\frac{1}{n}}

= ((y^{2} - 1)^{\frac{1}{3}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (y^{2} - 1)^{\frac{1}{3} \cdot 2}

\frac{1}{3} \cdot 2 = \frac{2}{3}

\frac{1}{3} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{3}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{3}

= (y^{2} - 1)^{\frac{2}{3}}

= (3 y^{2} - 1 + 1) ((y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1)

(3 y^{2} - 1 + 1) ((y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 y^{2} - 1 + 1 = 0 or (y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1 = 0

Solve

3 y^{2} - 1 + 1 = 0 : y = 0

3 y^{2} - 1 + 1 = 0

Move

1

to the right side

3 y^{2} - 1 + 1 = 0

Subtract

1

from both sides

3 y^{2} - 1 + 1 - 1 = 0 - 1

Simplify

3 y^{2} - 1 = - 1

3 y^{2} - 1 = - 1

Take both sides of the equation to the power of

3 : y^{2} - 1 = - 1

3 y^{2} - 1 = - 1

(3 y^{2} - 1)^{3} = (- 1)^{3}

Expand

(3 y^{2} - 1)^{3} : y^{2} - 1

(3 y^{2} - 1)^{3}

Apply radical rule:

n a = a^{\frac{1}{n}}

= ((y^{2} - 1)^{\frac{1}{3}})^{3}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (y^{2} - 1)^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

Multiply fractions:

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

= \frac{1 \cdot 3}{3}

Cancel the common factor:

3

= 1

= y^{2} - 1

Expand

(- 1)^{3} : - 1

(- 1)^{3}

Apply exponent rule:

(- a)^{n} = - a^{n},

n

is odd

(- 1)^{3} = - 1^{3}

= - 1^{3}

Apply rule

1^{a} = 1

= - 1

y^{2} - 1 = - 1

y^{2} - 1 = - 1

Solve

y^{2} - 1 = - 1 : y = 0

y^{2} - 1 = - 1

Move

1

to the right side

y^{2} - 1 = - 1

Add

1

to both sides

y^{2} - 1 + 1 = - 1 + 1

Simplify

y^{2} = 0

y^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

y = 0

y = 0

Verify Solutions:

y = 0

True

Check the solutions by plugging them into

3 y^{2} - 1 + 1 = 0

Remove the ones that don't agree with the equation.

Plug in

y = 0 :

True

3 0^{2} - 1 + 1 = 0

3 0^{2} - 1 + 1 = 0

3 0^{2} - 1 + 1

Apply rule

0^{a} = 0

0^{2} = 0

= 3 0 - 1 + 1

3 0 - 1 = - 1

3 0 - 1

Subtract the numbers:

0 - 1 = - 1

= 3 - 1

n - 1 = - 1,

n

is odd

3 - 1 = - 1

= - 1

= - 1 + 1

Add/Subtract the numbers:

- 1 + 1 = 0

= 0

0 = 0

True

The solution is

y = 0

Solve

(y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1 = 0 :

No Solution for

y \in R

(y^{2} - 1)^{\frac{2}{3}} - 3 y^{2} - 1 + 1 = 0

Use the following exponent property

: a^{\frac{n}{m}} = (m a)^{n}

(y^{2} - 1)^{\frac{2}{3}} = (3 y^{2} - 1)^{2}

(3 y^{2} - 1)^{2} - 3 y^{2} - 1 + 1 = 0

Rewrite the equation with

3 y^{2} - 1 = u

u^{2} - u + 1 = 0

Solve

u^{2} - u + 1 = 0 :

No Solution for

u \in R

u^{2} - u + 1 = 0

Discriminant

u^{2} - u + 1 = 0 : - 3

u^{2} - u + 1 = 0

For a quadratic equation of the form

a x^{2} + b x + c = 0

the discriminant is

b^{2} - 4 a c

For

a = 1, b = - 1, c = 1 : (- 1)^{2} - 4 \cdot 1 \cdot 1

(- 1)^{2} - 4 \cdot 1 \cdot 1

Expand

(- 1)^{2} - 4 \cdot 1 \cdot 1 : - 3

(- 1)^{2} - 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

- 3

Discriminant cannot be negative for

u \in R

The solution is

No Solution for u \in R

No Solution for y \in R

y = 0

Verify Solutions:

y = 0

True

Check the solutions by plugging them into

1 + 3 y^{2} - 1^{3} = 0

Remove the ones that don't agree with the equation.

Plug in

y = 0 :

True

1 + (3 0^{2} - 1)^{3} = 0

1 + (3 0^{2} - 1)^{3} = 0

1 + (3 0^{2} - 1)^{3}

Apply rule

0^{a} = 0

0^{2} = 0

= 1 + (3 0 - 1)^{3}

(3 0 - 1)^{3} = - 1

(3 0 - 1)^{3}

3 0 - 1 = - 1

3 0 - 1

Subtract the numbers:

0 - 1 = - 1

= 3 - 1

n - 1 = - 1,

n

is odd

3 - 1 = - 1

= - 1

= (- 1)^{3}

Apply exponent rule:

(- a)^{n} = - a^{n},

n

is odd

(- 1)^{3} = - 1^{3}

= - 1^{3}

Apply rule

1^{a} = 1

= - 1

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

0 = 0

True

The solution is

y = 0

y = 0

The intervals are defined around the zeroes:

y < 0, y = 0, y > 0

Combine intervals with domain

y < 0, y = 0, y > 0

Check the solutions by plugging them into

1 + 3 y^{2} - 1^{3} = y

Remove the ones that don't agree with the equation.

Plug

y < 0 : 1 + 3 y^{2} - 1^{3} \neq = y \Rightarrow

False

The solution is

y \geq 0

The solution is

sin (x y^{2}) = 3 y^{2} - 1 {y \geq 0}

Apply trig inverse properties

sin (x y^{2}) = 3 y^{2} - 1

General solutions for

sin (x y^{2}) = 3 y^{2} - 1

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn, x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn, x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

Solve

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn : x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn

Divide both sides by

y^{2}; y \neq = 0

x y^{2} = arcsin (3 y^{2} - 1) + 2 πn

Divide both sides by

y^{2}; y \neq = 0

\frac{x y ^{2}}{y ^{2}} = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

Simplify

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

Solve

x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn : x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

Divide both sides by

y^{2}; y \neq = 0

x y^{2} = π + arcsin (- 3 y^{2} - 1) + 2 πn

Divide both sides by

y^{2}; y \neq = 0

\frac{x y ^{2}}{y ^{2}} = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

Simplify

x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}; y \neq = 0

x = \frac{arcsin ( 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}, x = \frac{π}{y ^{2}} + \frac{arcsin ( - 3 y ^{2} - 1 )}{y ^{2}} + \frac{2 πn}{y ^{2}}

Graph

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