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sin^5(a)=16sin^5(a)-20sin^3(a)+5sin(a)

Solution

sin^{5} (a) = 16 sin^{5} (a) - 20 sin^{3} (a) + 5 sin (a)

Solution

a = 2 πn, a = π + 2 πn, a = - 0.61547 \dots + 2 πn, a = π + 0.61547 \dots + 2 πn, a = 0.61547 \dots + 2 πn, a = π - 0.61547 \dots + 2 πn, a = \frac{3 π}{2} + 2 πn, a = \frac{π}{2} + 2 πn

Degrees

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n, a = - 35.26438 \dots^{\circ} + 36 0^{\circ} n, a = 215.26438 \dots^{\circ} + 36 0^{\circ} n, a = 35.26438 \dots^{\circ} + 36 0^{\circ} n, a = 144.73561 \dots^{\circ} + 36 0^{\circ} n, a = 27 0^{\circ} + 36 0^{\circ} n, a = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{5} (a) = 16 sin^{5} (a) - 20 sin^{3} (a) + 5 sin (a)

Solve by substitution

sin^{5} (a) = 16 sin^{5} (a) - 20 sin^{3} (a) + 5 sin (a)

Let:

sin (a) = u

u^{5} = 16 u^{5} - 20 u^{3} + 5 u

u^{5} = 16 u^{5} - 20 u^{3} + 5 u : u = 0, u = - \frac{3}{3}, u = \frac{3}{3}, u = - 1, u = 1

u^{5} = 16 u^{5} - 20 u^{3} + 5 u

Switch sides

16 u^{5} - 20 u^{3} + 5 u = u^{5}

Move

u^{5}

to the left side

16 u^{5} - 20 u^{3} + 5 u = u^{5}

Subtract

u^{5}

from both sides

16 u^{5} - 20 u^{3} + 5 u - u^{5} = u^{5} - u^{5}

Simplify

15 u^{5} - 20 u^{3} + 5 u = 0

15 u^{5} - 20 u^{3} + 5 u = 0

Factor

15 u^{5} - 20 u^{3} + 5 u : 5 u (3 u + 1) (3 u - 1) (u + 1) (u - 1)

15 u^{5} - 20 u^{3} + 5 u

Factor out common term

5 u : 5 u (3 u^{4} - 4 u^{2} + 1)

15 u^{5} - 20 u^{3} + 5 u

Apply exponent rule:

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

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

= 15 u^{4} u - 20 u^{2} u + 5 u

Rewrite

20

5 \cdot 4

Rewrite

15

5 \cdot 3

= 5 \cdot 3 u^{4} u - 5 \cdot 4 u^{2} u + 5 u

Factor out common term

5 u

= 5 u (3 u^{4} - 4 u^{2} + 1)

= 5 u (3 u^{4} - 4 u^{2} + 1)

Factor

3 u^{4} - 4 u^{2} + 1 : (3 u + 1) (3 u - 1) (u + 1) (u - 1)

3 u^{4} - 4 u^{2} + 1

Let

u = u^{2}

= 3 u^{2} - 4 u + 1

Factor

3 u^{2} - 4 u + 1 : (3 u - 1) (u - 1)

3 u^{2} - 4 u + 1

Break the expression into groups

3 u^{2} - 4 u + 1

Definition

Factors of

3 : 1, 3

3

Divisors (Factors)

Find the Prime factors of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Add 1

1

The factors of

3

1, 3

Negative factors of

3 : - 1, - 3

Multiply the factors by

- 1

to get the negative factors

- 1, - 3

For every two factors such that

u * v = 3,

check if

u + v = - 4

Check

u = 1, v = 3 : u * v = 3, u + v = 4 \Rightarrow

FalseCheck

u = - 1, v = - 3 : u * v = 3, u + v = - 4 \Rightarrow

True

u = - 1, v = - 3

Group into

(a x^{2} + ux) + (vx + c)

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

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

Factor out

u

from

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

3 u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= 3 uu - u

Factor out common term

u

= u (3 u - 1)

Factor out

- 1

from

- 3 u + 1 : - (3 u - 1)

- 3 u + 1

Factor out common term

- 1

= - (3 u - 1)

= u (3 u - 1) - (3 u - 1)

Factor out common term

3 u - 1

= (3 u - 1) (u - 1)

= (3 u - 1) (u - 1)

Substitute back

u = u^{2}

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

Factor

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

3 u^{2} - 1

Rewrite

3 u^{2} - 1

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

3 u^{2} - 1

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(3)^{2} u^{2} = (3 u)^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

= (3 u + 1) (3 u - 1)

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

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (u + 1) (u - 1)

= (3 u + 1) (3 u - 1) (u + 1) (u - 1)

= 5 u (3 u + 1) (3 u - 1) (u + 1) (u - 1)

5 u (3 u + 1) (3 u - 1) (u + 1) (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 3 u + 1 = 0 or 3 u - 1 = 0 or u + 1 = 0 or u - 1 = 0

Solve

3 u + 1 = 0 : u = - \frac{3}{3}

3 u + 1 = 0

Move

1

to the right side

3 u + 1 = 0

Subtract

1

from both sides

3 u + 1 - 1 = 0 - 1

Simplify

3 u = - 1

3 u = - 1

Divide both sides by

3

3 u = - 1

Divide both sides by

3

\frac{3 u}{3} = \frac{- 1}{3}

Simplify

\frac{3 u}{3} = \frac{- 1}{3}

Simplify

\frac{3 u}{3} : u

\frac{3 u}{3}

Cancel the common factor:

3

= u

Simplify

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

\frac{- 1}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{3}

Rationalize

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

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

Solve

3 u - 1 = 0 : u = \frac{3}{3}

3 u - 1 = 0

Move

1

to the right side

3 u - 1 = 0

Add

1

to both sides

3 u - 1 + 1 = 0 + 1

Simplify

3 u = 1

3 u = 1

Divide both sides by

3

3 u = 1

Divide both sides by

3

\frac{3 u}{3} = \frac{1}{3}

Simplify

\frac{3 u}{3} = \frac{1}{3}

Simplify

\frac{3 u}{3} : u

\frac{3 u}{3}

Cancel the common factor:

3

= u

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The solutions are

u = 0, u = - \frac{3}{3}, u = \frac{3}{3}, u = - 1, u = 1

Substitute back

u = sin (a)

sin (a) = 0, sin (a) = - \frac{3}{3}, sin (a) = \frac{3}{3}, sin (a) = - 1, sin (a) = 1

sin (a) = 0, sin (a) = - \frac{3}{3}, sin (a) = \frac{3}{3}, sin (a) = - 1, sin (a) = 1

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

General solutions for

sin (a) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

sin (a) = - \frac{3}{3} : a = arcsin (- \frac{3}{3}) + 2 πn, a = π + arcsin (\frac{3}{3}) + 2 πn

sin (a) = - \frac{3}{3}

Apply trig inverse properties

sin (a) = - \frac{3}{3}

General solutions for

sin (a) = - \frac{3}{3}

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

a = arcsin (- \frac{3}{3}) + 2 πn, a = π + arcsin (\frac{3}{3}) + 2 πn

a = arcsin (- \frac{3}{3}) + 2 πn, a = π + arcsin (\frac{3}{3}) + 2 πn

sin (a) = \frac{3}{3} : a = arcsin (\frac{3}{3}) + 2 πn, a = π - arcsin (\frac{3}{3}) + 2 πn

sin (a) = \frac{3}{3}

Apply trig inverse properties

sin (a) = \frac{3}{3}

General solutions for

sin (a) = \frac{3}{3}

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

a = arcsin (\frac{3}{3}) + 2 πn, a = π - arcsin (\frac{3}{3}) + 2 πn

a = arcsin (\frac{3}{3}) + 2 πn, a = π - arcsin (\frac{3}{3}) + 2 πn

sin (a) = - 1 : a = \frac{3 π}{2} + 2 πn

sin (a) = - 1

General solutions for

sin (a) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = \frac{3 π}{2} + 2 πn

a = \frac{3 π}{2} + 2 πn

sin (a) = 1 : a = \frac{π}{2} + 2 πn

sin (a) = 1

General solutions for

sin (a) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = \frac{π}{2} + 2 πn

a = \frac{π}{2} + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn, a = arcsin (- \frac{3}{3}) + 2 πn, a = π + arcsin (\frac{3}{3}) + 2 πn, a = arcsin (\frac{3}{3}) + 2 πn, a = π - arcsin (\frac{3}{3}) + 2 πn, a = \frac{3 π}{2} + 2 πn, a = \frac{π}{2} + 2 πn

Show solutions in decimal form

a = 2 πn, a = π + 2 πn, a = - 0.61547 \dots + 2 πn, a = π + 0.61547 \dots + 2 πn, a = 0.61547 \dots + 2 πn, a = π - 0.61547 \dots + 2 πn, a = \frac{3 π}{2} + 2 πn, a = \frac{π}{2} + 2 πn

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