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

The general solution for sin^{sin(x)}(x)=2 is No Solution for x\in\mathbb{R}

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Popular Trigonometry >

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

Solution

sin^{s i n (x)} (x) = 2

Solution

No Solution for x \in R

Solution steps

sin^{s i n (x)} (x) = 2

Solve by substitution

sin^{s i n (x)} (x) = 2

Let:

sin (x) = u

u^{u} = 2

u^{u} = 2 : u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

u^{u} = 2

Prepare

u^{u} = 2

for Lambert form:

u e^{- \frac{l n ( 2 )}{u}} = 1

u^{u} = 2

x e^{x} = a

is equation in Lambert form

Take both sides of the equation to the power of

\frac{1}{u}

(u^{u})^{\frac{1}{u}} = 2^{\frac{1}{u}}

Simplify

(u^{u})^{\frac{1}{u}} : u

(u^{u})^{\frac{1}{u}}

Apply exponent rule:

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

assuming

a \geq 0

= u^{u \frac{1}{u}}

u \frac{1}{u} = 1

u \frac{1}{u}

Multiply fractions:

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

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

Cancel the common factor:

u

= 1

= u

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

Multiply both sides by

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

Add similar elements:

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

= 2^{0}

Apply rule

a^{0} = 1, a \neq = 0

= 1

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

Apply exponent rules

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

Convert to base

e : u e^{l n (2) (- \frac{1}{u})} = 1

Apply exponent rule:

a = b^{l o g_{b} (a)}

2^{- \frac{1}{u}} = (e^{l n (2)})^{- \frac{1}{u}}

u (e^{l n (2)})^{- \frac{1}{u}} = 1

Apply exponent rule:

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

(e^{l n (2)})^{- \frac{1}{u}} = e^{l n (2) (- \frac{1}{u})}

u e^{l n (2) (- \frac{1}{u})} = 1

u e^{l n (2) (- \frac{1}{u})} = 1

Simplify

u e^{- \frac{l n ( 2 )}{u}} = 1

u e^{- \frac{l n ( 2 )}{u}} = 1

Rewrite the equation with

\frac{ln ( 2 )}{u} = v

and

u = \frac{ln ( 2 )}{v}

(\frac{ln ( 2 )}{v}) e^{- v} = 1

Rewrite

(\frac{ln ( 2 )}{v}) e^{- v} = 1

in Lambert form:

v e^{v} = ln (2)

(\frac{ln ( 2 )}{v}) e^{- v} = 1

x e^{x} = a

is equation in Lambert form

Multiply both sides by

v

\frac{ln ( 2 )}{v} e^{- v} v = 1 \cdot v

Simplify

ln (2) e^{- v} = v

Multiply both sides by

e^{v}

ln (2) e^{- v} e^{v} = v e^{v}

Simplify

ln (2) e^{- v} e^{v} : ln (2)

ln (2) e^{- v} e^{v}

Apply exponent rule:

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

e^{- v} e^{v} = e^{- v + v}

= ln (2) e^{- v + v}

Add similar elements:

- v + v = 0

= ln (2) e^{0}

Apply rule

a^{0} = 1, a \neq = 0

= 1 \cdot ln (2)

Multiply:

ln (2) \cdot 1 = ln (2)

= ln (2)

ln (2) = v e^{v}

Switch sides

v e^{v} = ln (2)

Solve

v e^{v} = ln (2) : v = W_{0} (ln (2))

v e^{v} = ln (2)

Solution for

x e^{x} = a

where

a \geq 0

is principal branch of Lambert

W

function:

x = W_{0} (a)

v = W_{0} (ln (2))

Verify Solutions:

v = W_{0} (ln (2))

True

Check the solutions by plugging them into

(\frac{ln ( 2 )}{v}) e^{- v} = 1

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

Plug in

v = W_{0} (ln (2)) :

True

(\frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}) e^{- W_{0} (l n (2))} = 1

(\frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}) e^{- W_{0} (l n (2))} = \frac{e ^{- W_{0} (l n (2))} ln ( 2 )}{W _{0} ( ln ( 2 ) )}

(\frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}) e^{- W_{0} (l n (2))}

Remove parentheses:

(a) = a

= \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )} e^{- W_{0} (l n (2))}

Multiply fractions:

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

= \frac{ln ( 2 ) e ^{- W_{0} (l n (2))}}{W _{0} ( ln ( 2 ) )}

\frac{e ^{- W_{0} (l n (2))} ln ( 2 )}{W _{0} ( ln ( 2 ) )} = 1

True

The solution is

v = W_{0} (ln (2))

Substitute back

v = \frac{ln ( 2 )}{u},

solve for

u

Solve

\frac{ln ( 2 )}{u} = W_{0} (ln (2)) : u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

\frac{ln ( 2 )}{u} = W_{0} (ln (2))

Multiply both sides by

u

\frac{ln ( 2 )}{u} = W_{0} (ln (2))

Multiply both sides by

u

\frac{ln ( 2 )}{u} u = W_{0} (ln (2)) u

Simplify

ln (2) = W_{0} (ln (2)) u

ln (2) = W_{0} (ln (2)) u

Switch sides

W_{0} (ln (2)) u = ln (2)

Divide both sides by

W_{0} (ln (2))

W_{0} (ln (2)) u = ln (2)

Divide both sides by

W_{0} (ln (2))

\frac{W _{0} ( ln ( 2 ) ) u}{W _{0} ( ln ( 2 ) )} = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

Simplify

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{ln ( 2 )}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

Substitute back

u = sin (x)

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )} :

No Solution

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sin^{sin(x)}(x)=2 ?
The general solution for sin^{sin(x)}(x)=2 is No Solution for x\in\mathbb{R}