What is the general solution for 8(1-cos(θ))=sin^2(θ) ?

The general solution for 8(1-cos(θ))=sin^2(θ) is θ=2pin

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8(1-cos(θ))=sin^2(θ)

Solution

8 (1 - cos (θ)) = sin^{2} (θ)

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

8 (1 - cos (θ)) = sin^{2} (θ)

Subtract

sin^{2} (θ)

from both sides

8 (1 - cos (θ)) - sin^{2} (θ) = 0

Rewrite using trig identities

- sin^{2} (θ) + (1 - cos (θ)) \cdot 8

Use the Pythagorean identity:

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

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

= - (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 8

Simplify

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 8 : cos^{2} (θ) - 8 cos (θ) + 7

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 8

= - (1 - cos^{2} (θ)) + 8 (1 - cos (θ))

- (1 - cos^{2} (θ)) : - 1 + cos^{2} (θ)

- (1 - cos^{2} (θ))

Distribute parentheses

= - (1) - (- cos^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (θ)

= - 1 + cos^{2} (θ) + (1 - cos (θ)) \cdot 8

Expand

8 (1 - cos (θ)) : 8 - 8 cos (θ)

8 (1 - cos (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 1, c = cos (θ)

= 8 \cdot 1 - 8 cos (θ)

Multiply the numbers:

8 \cdot 1 = 8

= 8 - 8 cos (θ)

= - 1 + cos^{2} (θ) + 8 - 8 cos (θ)

Simplify

- 1 + cos^{2} (θ) + 8 - 8 cos (θ) : cos^{2} (θ) - 8 cos (θ) + 7

- 1 + cos^{2} (θ) + 8 - 8 cos (θ)

Group like terms

= cos^{2} (θ) - 8 cos (θ) - 1 + 8

Add/Subtract the numbers:

- 1 + 8 = 7

= cos^{2} (θ) - 8 cos (θ) + 7

= cos^{2} (θ) - 8 cos (θ) + 7

= cos^{2} (θ) - 8 cos (θ) + 7

7 + cos^{2} (θ) - 8 cos (θ) = 0

Solve by substitution

7 + cos^{2} (θ) - 8 cos (θ) = 0

Let:

cos (θ) = u

7 + u^{2} - 8 u = 0

7 + u^{2} - 8 u = 0 : u = 7, u = 1

7 + u^{2} - 8 u = 0

Write in the standard form

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

u^{2} - 8 u + 7 = 0

Solve with the quadratic formula

u^{2} - 8 u + 7 = 0

Quadratic Equation Formula:

For

a = 1, b = - 8, c = 7

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 7 = 6

(- 8)^{2} - 4 \cdot 1 \cdot 7

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 1 \cdot 7

Multiply the numbers:

4 \cdot 1 \cdot 7 = 28

= 8^{2} - 28

8^{2} = 64

= 64 - 28

Subtract the numbers:

64 - 28 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 8 ) \pm 6}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 6}{2 \cdot 1}, u_{2} = \frac{- ( - 8 ) - 6}{2 \cdot 1}

u = \frac{- ( - 8 ) + 6}{2 \cdot 1} : 7

\frac{- ( - 8 ) + 6}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{8 + 6}{2 \cdot 1}

Add the numbers:

8 + 6 = 14

= \frac{14}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{14}{2}

Divide the numbers:

\frac{14}{2} = 7

= 7

u = \frac{- ( - 8 ) - 6}{2 \cdot 1} : 1

\frac{- ( - 8 ) - 6}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{8 - 6}{2 \cdot 1}

Subtract the numbers:

8 - 6 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 7, u = 1

Substitute back

u = cos (θ)

cos (θ) = 7, cos (θ) = 1

cos (θ) = 7, cos (θ) = 1

cos (θ) = 7 :

No Solution

cos (θ) = 7

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 1

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

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 8(1-cos(θ))=sin^2(θ) ?
The general solution for 8(1-cos(θ))=sin^2(θ) is θ=2pin