What is the general solution for 4sin(x)-sin^3(x)-1=0 ?

The general solution for 4sin(x)-sin^3(x)-1=0 is x=0.25691…+2pin,x=pi-0.25691…+2pin

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4sin(x)-sin^3(x)-1=0

Solution

4 sin (x) - sin^{3} (x) - 1 = 0

Solution

x = 0.25691 \dots + 2 πn, x = π - 0.25691 \dots + 2 πn

Degrees

x = 14.72036 \dots^{\circ} + 36 0^{\circ} n, x = 165.27963 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (x) - sin^{3} (x) - 1 = 0

Solve by substitution

4 sin (x) - sin^{3} (x) - 1 = 0

Let:

sin (x) = u

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

4 u - u^{3} - 1 = 0 : u \approx 0.25410 \dots, u \approx 1.86080 \dots, u \approx - 2.11490 \dots

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{3} + 4 u - 1 = 0

Find one solution for

- u^{3} + 4 u - 1 = 0

using Newton-Raphson:

u \approx 0.25410 \dots

- u^{3} + 4 u - 1 = 0

Newton-Raphson Approximation Definition

f (u) = - u^{3} + 4 u - 1

Find

f^{^{'}} (u) : - 3 u^{2} + 4

\frac{d}{d u} (- u^{3} + 4 u - 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (u^{3}) + \frac{d}{d u} (4 u) - \frac{d}{d u} (1)

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

Simplify

= 3 u^{2}

\frac{d}{d u} (4 u) = 4

\frac{d}{d u} (4 u)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d u}{d u}

Apply the common derivative:

\frac{d u}{d u} = 1

= 4 \cdot 1

Simplify

= 4

\frac{d}{d u} (1) = 0

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

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= - 3 u^{2} + 4 - 0

Simplify

= - 3 u^{2} + 4

Let

u_{0} = 0

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = 0.25 : Δ u_{1} = 0.25

f (u_{0}) = - 0^{3} + 4 \cdot 0 - 1 = - 1

f^{^{'}} (u_{0}) = - 3 \cdot 0^{2} + 4 = 4

u_{1} = 0.25

Δ u_{1} = ∣ 0.25 - 0 ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = 0.25409 \dots : Δ u_{2} = 0.00409 \dots

f (u_{1}) = - 0.2 5^{3} + 4 \cdot 0.25 - 1 = - 0.015625

f^{^{'}} (u_{1}) = - 3 \cdot 0.2 5^{2} + 4 = 3.8125

u_{2} = 0.25409 \dots

Δ u_{2} = ∣ 0.25409 \dots - 0.25 ∣ = 0.00409 \dots

Δ u_{2} = 0.00409 \dots

u_{3} = 0.25410 \dots : Δ u_{3} = 3.32771 E - 6

f (u_{2}) = - 0.25409 \dots^{3} + 4 \cdot 0.25409 \dots - 1 = - 0.00001 \dots

f^{^{'}} (u_{2}) = - 3 \cdot 0.25409 \dots^{2} + 4 = 3.80630 \dots

u_{3} = 0.25410 \dots

Δ u_{3} = ∣ 0.25410 \dots - 0.25409 \dots ∣ = 3.32771 E - 6

Δ u_{3} = 3.32771 E - 6

u_{4} = 0.25410 \dots : Δ u_{4} = 2.21776 E - 12

f (u_{3}) = - 0.25410 \dots^{3} + 4 \cdot 0.25410 \dots - 1 = - 8.44147 E - 12

f^{^{'}} (u_{3}) = - 3 \cdot 0.25410 \dots^{2} + 4 = 3.80629 \dots

u_{4} = 0.25410 \dots

Δ u_{4} = ∣ 0.25410 \dots - 0.25410 \dots ∣ = 2.21776 E - 12

Δ u_{4} = 2.21776 E - 12

u \approx 0.25410 \dots

Apply long division:

\frac{- u ^{3} + 4 u - 1}{u - 0.25410 \dots} = - u^{2} - 0.25410 \dots u + 3.93543 \dots

- u^{2} - 0.25410 \dots u + 3.93543 \dots \approx 0

Find one solution for

- u^{2} - 0.25410 \dots u + 3.93543 \dots = 0

using Newton-Raphson:

u \approx 1.86080 \dots

- u^{2} - 0.25410 \dots u + 3.93543 \dots = 0

Newton-Raphson Approximation Definition

f (u) = - u^{2} - 0.25410 \dots u + 3.93543 \dots

Find

f^{^{'}} (u) : - 2 u - 0.25410 \dots

\frac{d}{d u} (- u^{2} - 0.25410 \dots u + 3.93543 \dots)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (u^{2}) - \frac{d}{d u} (0.25410 \dots u) + \frac{d}{d u} (3.93543 \dots)

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

\frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplify

= 2 u

\frac{d}{d u} (0.25410 \dots u) = 0.25410 \dots

\frac{d}{d u} (0.25410 \dots u)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.25410 \dots \frac{d u}{d u}

Apply the common derivative:

\frac{d u}{d u} = 1

= 0.25410 \dots \cdot 1

Simplify

= 0.25410 \dots

\frac{d}{d u} (3.93543 \dots) = 0

\frac{d}{d u} (3.93543 \dots)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= - 2 u - 0.25410 \dots + 0

Simplify

= - 2 u - 0.25410 \dots

Let

u_{0} = 5

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = 2.82183 \dots : Δ u_{1} = 2.17816 \dots

f (u_{0}) = - 5^{2} - 0.25410 \dots \cdot 5 + 3.93543 \dots = - 22.33507 \dots

f^{^{'}} (u_{0}) = - 2 \cdot 5 - 0.25410 \dots = - 10.25410 \dots

u_{1} = 2.82183 \dots

Δ u_{1} = ∣ 2.82183 \dots - 5 ∣ = 2.17816 \dots

Δ u_{1} = 2.17816 \dots

u_{2} = 2.01740 \dots : Δ u_{2} = 0.80443 \dots

f (u_{1}) = - 2.82183 \dots^{2} - 0.25410 \dots \cdot 2.82183 \dots + 3.93543 \dots = - 4.74438 \dots

f^{^{'}} (u_{1}) = - 2 \cdot 2.82183 \dots - 0.25410 \dots = - 5.89778 \dots

u_{2} = 2.01740 \dots

Δ u_{2} = ∣ 2.01740 \dots - 2.82183 \dots ∣ = 0.80443 \dots

Δ u_{2} = 0.80443 \dots

u_{3} = 1.86652 \dots : Δ u_{3} = 0.15088 \dots

f (u_{2}) = - 2.01740 \dots^{2} - 0.25410 \dots \cdot 2.01740 \dots + 3.93543 \dots = - 0.64711 \dots

f^{^{'}} (u_{2}) = - 2 \cdot 2.01740 \dots - 0.25410 \dots = - 4.28891 \dots

u_{3} = 1.86652 \dots

Δ u_{3} = ∣ 1.86652 \dots - 2.01740 \dots ∣ = 0.15088 \dots

Δ u_{3} = 0.15088 \dots

u_{4} = 1.86081 \dots : Δ u_{4} = 0.00570 \dots

f (u_{3}) = - 1.86652 \dots^{2} - 0.25410 \dots \cdot 1.86652 \dots + 3.93543 \dots = - 0.02276 \dots

f^{^{'}} (u_{3}) = - 2 \cdot 1.86652 \dots - 0.25410 \dots = - 3.98714 \dots

u_{4} = 1.86081 \dots

Δ u_{4} = ∣ 1.86081 \dots - 1.86652 \dots ∣ = 0.00570 \dots

Δ u_{4} = 0.00570 \dots

u_{5} = 1.86080 \dots : Δ u_{5} = 8.1997 E - 6

f (u_{4}) = - 1.86081 \dots^{2} - 0.25410 \dots \cdot 1.86081 \dots + 3.93543 \dots = - 0.00003 \dots

f^{^{'}} (u_{4}) = - 2 \cdot 1.86081 \dots - 0.25410 \dots = - 3.97572 \dots

u_{5} = 1.86080 \dots

Δ u_{5} = ∣ 1.86080 \dots - 1.86081 \dots ∣ = 8.1997 E - 6

Δ u_{5} = 8.1997 E - 6

u_{6} = 1.86080 \dots : Δ u_{6} = 1.69115 E - 11

f (u_{5}) = - 1.86080 \dots^{2} - 0.25410 \dots \cdot 1.86080 \dots + 3.93543 \dots = - 6.72351 E - 11

f^{^{'}} (u_{5}) = - 2 \cdot 1.86080 \dots - 0.25410 \dots = - 3.97571 \dots

u_{6} = 1.86080 \dots

Δ u_{6} = ∣ 1.86080 \dots - 1.86080 \dots ∣ = 1.69115 E - 11

Δ u_{6} = 1.69115 E - 11

u \approx 1.86080 \dots

Apply long division:

\frac{- u ^{2} - 0.25410 \dots u + 3.93543 \dots}{u - 1.86080 \dots} = - u - 2.11490 \dots

- u - 2.11490 \dots \approx 0

u \approx - 2.11490 \dots

The solutions are

u \approx 0.25410 \dots, u \approx 1.86080 \dots, u \approx - 2.11490 \dots

Substitute back

u = sin (x)

sin (x) \approx 0.25410 \dots, sin (x) \approx 1.86080 \dots, sin (x) \approx - 2.11490 \dots

sin (x) \approx 0.25410 \dots, sin (x) \approx 1.86080 \dots, sin (x) \approx - 2.11490 \dots

sin (x) = 0.25410 \dots : x = arcsin (0.25410 \dots) + 2 πn, x = π - arcsin (0.25410 \dots) + 2 πn

sin (x) = 0.25410 \dots

Apply trig inverse properties

sin (x) = 0.25410 \dots

General solutions for

sin (x) = 0.25410 \dots

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

x = arcsin (0.25410 \dots) + 2 πn, x = π - arcsin (0.25410 \dots) + 2 πn

x = arcsin (0.25410 \dots) + 2 πn, x = π - arcsin (0.25410 \dots) + 2 πn

sin (x) = 1.86080 \dots :

No Solution

sin (x) = 1.86080 \dots

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - 2.11490 \dots :

No Solution

sin (x) = - 2.11490 \dots

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (0.25410 \dots) + 2 πn, x = π - arcsin (0.25410 \dots) + 2 πn

Show solutions in decimal form

x = 0.25691 \dots + 2 πn, x = π - 0.25691 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(x)-sin^3(x)-1=0 ?
The general solution for 4sin(x)-sin^3(x)-1=0 is x=0.25691…+2pin,x=pi-0.25691…+2pin