What is the general solution for cos(x)=(-3)/(5sin^2(x)) ?

The general solution for cos(x)=(-3)/(5sin^2(x)) is No Solution for x\in\mathbb{R}

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cos(x)=(-3)/(5sin^2(x))

Solution

cos (x) = \frac{- 3}{5 sin ^{2} ( x )}

Solution

No Solution for x \in R

Solution steps

cos (x) = \frac{- 3}{5 sin ^{2} ( x )}

Subtract

\frac{- 3}{5 sin ^{2} ( x )}

from both sides

cos (x) + \frac{3}{5 sin ^{2} ( x )} = 0

Simplify

cos (x) + \frac{3}{5 sin ^{2} ( x )} : \frac{5 sin ^{2} ( x ) cos ( x ) + 3}{5 sin ^{2} ( x )}

cos (x) + \frac{3}{5 sin ^{2} ( x )}

Convert element to fraction:

cos (x) = \frac{cos ( x ) 5 sin ^{2} ( x )}{5 sin ^{2} ( x )}

= \frac{cos ( x ) \cdot 5 sin ^{2} ( x )}{5 sin ^{2} ( x )} + \frac{3}{5 sin ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( x ) \cdot 5 sin ^{2} ( x ) + 3}{5 sin ^{2} ( x )}

\frac{5 sin ^{2} ( x ) cos ( x ) + 3}{5 sin ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

5 sin^{2} (x) cos (x) + 3 = 0

Rewrite using trig identities

3 + 5 cos (x) sin^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 3 + 5 cos (x) (1 - cos^{2} (x))

3 + (1 - cos^{2} (x)) \cdot 5 cos (x) = 0

Solve by substitution

3 + (1 - cos^{2} (x)) \cdot 5 cos (x) = 0

Let:

cos (x) = u

3 + (1 - u^{2}) \cdot 5 u = 0

3 + (1 - u^{2}) \cdot 5 u = 0 : u \approx 1.22119 \dots

3 + (1 - u^{2}) \cdot 5 u = 0

Expand

3 + (1 - u^{2}) \cdot 5 u : 3 + 5 u - 5 u^{3}

3 + (1 - u^{2}) \cdot 5 u

= 3 + 5 u (1 - u^{2})

Expand

5 u (1 - u^{2}) : 5 u - 5 u^{3}

5 u (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 5 u, b = 1, c = u^{2}

= 5 u \cdot 1 - 5 u u^{2}

= 5 \cdot 1 \cdot u - 5 u^{2} u

Simplify

5 \cdot 1 \cdot u - 5 u^{2} u : 5 u - 5 u^{3}

5 \cdot 1 \cdot u - 5 u^{2} u

5 \cdot 1 \cdot u = 5 u

5 \cdot 1 \cdot u

Multiply the numbers:

5 \cdot 1 = 5

= 5 u

5 u^{2} u = 5 u^{3}

5 u^{2} u

Apply exponent rule:

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

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

= 5 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 5 u^{3}

= 5 u - 5 u^{3}

= 5 u - 5 u^{3}

= 3 + 5 u - 5 u^{3}

3 + 5 u - 5 u^{3} = 0

Write in the standard form

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

- 5 u^{3} + 5 u + 3 = 0

Find one solution for

- 5 u^{3} + 5 u + 3 = 0

using Newton-Raphson:

u \approx 1.22119 \dots

- 5 u^{3} + 5 u + 3 = 0

Newton-Raphson Approximation Definition

f (u) = - 5 u^{3} + 5 u + 3

Find

f^{^{'}} (u) : - 15 u^{2} + 5

\frac{d}{d u} (- 5 u^{3} + 5 u + 3)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d u} (5 u^{3}) + \frac{d}{d u} (5 u) + \frac{d}{d u} (3)

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

Take the constant out:

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

= 5 \frac{d}{d u} (u^{3})

Apply the Power Rule:

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

= 5 \cdot 3 u^{3 - 1}

Simplify

= 15 u^{2}

\frac{d}{d u} (5 u) = 5

\frac{d}{d u} (5 u)

Take the constant out:

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

= 5 \frac{d u}{d u}

Apply the common derivative:

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

= 5 \cdot 1

Simplify

= 5

\frac{d}{d u} (3) = 0

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

Derivative of a constant:

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

= 0

= - 15 u^{2} + 5 + 0

Simplify

= - 15 u^{2} + 5

Let

u_{0} = 1

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = 1.3 : Δ u_{1} = 0.3

f (u_{0}) = - 5 \cdot 1^{3} + 5 \cdot 1 + 3 = 3

f^{^{'}} (u_{0}) = - 15 \cdot 1^{2} + 5 = - 10

u_{1} = 1.3

Δ u_{1} = ∣ 1.3 - 1 ∣ = 0.3

Δ u_{1} = 0.3

u_{2} = 1.22702 \dots : Δ u_{2} = 0.07297 \dots

f (u_{1}) = - 5 \cdot 1. 3^{3} + 5 \cdot 1.3 + 3 = - 1.485

f^{^{'}} (u_{1}) = - 15 \cdot 1. 3^{2} + 5 = - 20.35

u_{2} = 1.22702 \dots

Δ u_{2} = ∣ 1.22702 \dots - 1.3 ∣ = 0.07297 \dots

Δ u_{2} = 0.07297 \dots

u_{3} = 1.22123 \dots : Δ u_{3} = 0.00579 \dots

f (u_{2}) = - 5 \cdot 1.22702 \dots^{3} + 5 \cdot 1.22702 \dots + 3 = - 0.10189 \dots

f^{^{'}} (u_{2}) = - 15 \cdot 1.22702 \dots^{2} + 5 = - 17.58392 \dots

u_{3} = 1.22123 \dots

Δ u_{3} = ∣ 1.22123 \dots - 1.22702 \dots ∣ = 0.00579 \dots

Δ u_{3} = 0.00579 \dots

u_{4} = 1.22119 \dots : Δ u_{4} = 0.00003 \dots

f (u_{3}) = - 5 \cdot 1.22123 \dots^{3} + 5 \cdot 1.22123 \dots + 3 = - 0.00061 \dots

f^{^{'}} (u_{3}) = - 15 \cdot 1.22123 \dots^{2} + 5 = - 17.37112 \dots

u_{4} = 1.22119 \dots

Δ u_{4} = ∣ 1.22119 \dots - 1.22123 \dots ∣ = 0.00003 \dots

Δ u_{4} = 0.00003 \dots

u_{5} = 1.22119 \dots : Δ u_{5} = 1.33081 E - 9

f (u_{4}) = - 5 \cdot 1.22119 \dots^{3} + 5 \cdot 1.22119 \dots + 3 = - 2.31159 E - 8

f^{^{'}} (u_{4}) = - 15 \cdot 1.22119 \dots^{2} + 5 = - 17.36982 \dots

u_{5} = 1.22119 \dots

Δ u_{5} = ∣ 1.22119 \dots - 1.22119 \dots ∣ = 1.33081 E - 9

Δ u_{5} = 1.33081 E - 9

u \approx 1.22119 \dots

Apply long division:

\frac{- 5 u ^{3} + 5 u + 3}{u - 1.22119 \dots} = - 5 u^{2} - 6.10598 \dots u - 2.45660 \dots

- 5 u^{2} - 6.10598 \dots u - 2.45660 \dots \approx 0

Find one solution for

- 5 u^{2} - 6.10598 \dots u - 2.45660 \dots = 0

using Newton-Raphson:

No Solution for

u \in R

- 5 u^{2} - 6.10598 \dots u - 2.45660 \dots = 0

Newton-Raphson Approximation Definition

f (u) = - 5 u^{2} - 6.10598 \dots u - 2.45660 \dots

Find

f^{^{'}} (u) : - 10 u - 6.10598 \dots

\frac{d}{d u} (- 5 u^{2} - 6.10598 \dots u - 2.45660 \dots)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d u} (5 u^{2}) - \frac{d}{d u} (6.10598 \dots u) - \frac{d}{d u} (2.45660 \dots)

\frac{d}{d u} (5 u^{2}) = 10 u

\frac{d}{d u} (5 u^{2})

Take the constant out:

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

= 5 \frac{d}{d u} (u^{2})

Apply the Power Rule:

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

= 5 \cdot 2 u^{2 - 1}

Simplify

= 10 u

\frac{d}{d u} (6.10598 \dots u) = 6.10598 \dots

\frac{d}{d u} (6.10598 \dots u)

Take the constant out:

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

= 6.10598 \dots \frac{d u}{d u}

Apply the common derivative:

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

= 6.10598 \dots \cdot 1

Simplify

= 6.10598 \dots

\frac{d}{d u} (2.45660 \dots) = 0

\frac{d}{d u} (2.45660 \dots)

Derivative of a constant:

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

= 0

= - 10 u - 6.10598 \dots - 0

Simplify

= - 10 u - 6.10598 \dots

Let

u_{0} = 0

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = - 0.40232 \dots : Δ u_{1} = 0.40232 \dots

f (u_{0}) = - 5 \cdot 0^{2} - 6.10598 \dots \cdot 0 - 2.45660 \dots = - 2.45660 \dots

f^{^{'}} (u_{0}) = - 10 \cdot 0 - 6.10598 \dots = - 6.10598 \dots

u_{1} = - 0.40232 \dots

Δ u_{1} = ∣ - 0.40232 \dots - 0 ∣ = 0.40232 \dots

Δ u_{1} = 0.40232 \dots

u_{2} = - 0.79092 \dots : Δ u_{2} = 0.38859 \dots

f (u_{1}) = - 5 (- 0.40232 \dots)^{2} - 6.10598 \dots (- 0.40232 \dots) - 2.45660 \dots = - 0.80933 \dots

f^{^{'}} (u_{1}) = - 10 (- 0.40232 \dots) - 6.10598 \dots = - 2.08270 \dots

u_{2} = - 0.79092 \dots

Δ u_{2} = ∣ - 0.79092 \dots - (- 0.40232 \dots) ∣ = 0.38859 \dots

Δ u_{2} = 0.38859 \dots

u_{3} = - 0.37222 \dots : Δ u_{3} = 0.41870 \dots

f (u_{2}) = - 5 (- 0.79092 \dots)^{2} - 6.10598 \dots (- 0.79092 \dots) - 2.45660 \dots = - 0.75504 \dots

f^{^{'}} (u_{2}) = - 10 (- 0.79092 \dots) - 6.10598 \dots = 1.80328 \dots

u_{3} = - 0.37222 \dots

Δ u_{3} = ∣ - 0.37222 \dots - (- 0.79092 \dots) ∣ = 0.41870 \dots

Δ u_{3} = 0.41870 \dots

u_{4} = - 0.73994 \dots : Δ u_{4} = 0.36772 \dots

f (u_{3}) = - 5 (- 0.37222 \dots)^{2} - 6.10598 \dots (- 0.37222 \dots) - 2.45660 \dots = - 0.87657 \dots

f^{^{'}} (u_{3}) = - 10 (- 0.37222 \dots) - 6.10598 \dots = - 2.38377 \dots

u_{4} = - 0.73994 \dots

Δ u_{4} = ∣ - 0.73994 \dots - (- 0.37222 \dots) ∣ = 0.36772 \dots

Δ u_{4} = 0.36772 \dots

u_{5} = - 0.21724 \dots : Δ u_{5} = 0.52270 \dots

f (u_{4}) = - 5 (- 0.73994 \dots)^{2} - 6.10598 \dots (- 0.73994 \dots) - 2.45660 \dots = - 0.67611 \dots

f^{^{'}} (u_{4}) = - 10 (- 0.73994 \dots) - 6.10598 \dots = 1.29348 \dots

u_{5} = - 0.21724 \dots

Δ u_{5} = ∣ - 0.21724 \dots - (- 0.73994 \dots) ∣ = 0.52270 \dots

Δ u_{5} = 0.52270 \dots

u_{6} = - 0.56453 \dots : Δ u_{6} = 0.34729 \dots

f (u_{5}) = - 5 (- 0.21724 \dots)^{2} - 6.10598 \dots (- 0.21724 \dots) - 2.45660 \dots = - 1.36610 \dots

f^{^{'}} (u_{5}) = - 10 (- 0.21724 \dots) - 6.10598 \dots = - 3.93356 \dots

u_{6} = - 0.56453 \dots

Δ u_{6} = ∣ - 0.56453 \dots - (- 0.21724 \dots) ∣ = 0.34729 \dots

Δ u_{6} = 0.34729 \dots

u_{7} = - 1.87375 \dots : Δ u_{7} = 1.30922 \dots

f (u_{6}) = - 5 (- 0.56453 \dots)^{2} - 6.10598 \dots (- 0.56453 \dots) - 2.45660 \dots = - 0.60306 \dots

f^{^{'}} (u_{6}) = - 10 (- 0.56453 \dots) - 6.10598 \dots = - 0.46062 \dots

u_{7} = - 1.87375 \dots

Δ u_{7} = ∣ - 1.87375 \dots - (- 0.56453 \dots) ∣ = 1.30922 \dots

Δ u_{7} = 1.30922 \dots

u_{8} = - 1.19527 \dots : Δ u_{8} = 0.67848 \dots

f (u_{7}) = - 5 (- 1.87375 \dots)^{2} - 6.10598 \dots (- 1.87375 \dots) - 2.45660 \dots = - 8.57033 \dots

f^{^{'}} (u_{7}) = - 10 (- 1.87375 \dots) - 6.10598 \dots = 12.63161 \dots

u_{8} = - 1.19527 \dots

Δ u_{8} = ∣ - 1.19527 \dots - (- 1.87375 \dots) ∣ = 0.67848 \dots

Δ u_{8} = 0.67848 \dots

u_{9} = - 0.80160 \dots : Δ u_{9} = 0.39366 \dots

f (u_{8}) = - 5 (- 1.19527 \dots)^{2} - 6.10598 \dots (- 1.19527 \dots) - 2.45660 \dots = - 2.30169 \dots

f^{^{'}} (u_{8}) = - 10 (- 1.19527 \dots) - 6.10598 \dots = 5.84678 \dots

u_{9} = - 0.80160 \dots

Δ u_{9} = ∣ - 0.80160 \dots - (- 1.19527 \dots) ∣ = 0.39366 \dots

Δ u_{9} = 0.39366 \dots

u_{10} = - 0.39593 \dots : Δ u_{10} = 0.40567 \dots

f (u_{9}) = - 5 (- 0.80160 \dots)^{2} - 6.10598 \dots (- 0.80160 \dots) - 2.45660 \dots = - 0.77487 \dots

f^{^{'}} (u_{9}) = - 10 (- 0.80160 \dots) - 6.10598 \dots = 1.91008 \dots

u_{10} = - 0.39593 \dots

Δ u_{10} = ∣ - 0.39593 \dots - (- 0.80160 \dots) ∣ = 0.40567 \dots

Δ u_{10} = 0.40567 \dots

Cannot find solution

The solution is

u \approx 1.22119 \dots

Substitute back

u = cos (x)

cos (x) \approx 1.22119 \dots

cos (x) \approx 1.22119 \dots

cos (x) = 1.22119 \dots :

No Solution

cos (x) = 1.22119 \dots

- 1 \leq cos (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 cos(x)=(-3)/(5sin^2(x)) ?
The general solution for cos(x)=(-3)/(5sin^2(x)) is No Solution for x\in\mathbb{R}