What is the general solution for 6(1-cos(θ))=sin^2(θ),0<= θ<= 2pi ?

Q: What is the general solution for 6(1-cos(θ))=sin^2(θ),0<= θ<= 2pi ?

The general solution for 6(1-cos(θ))=sin^2(θ),0<= θ<= 2pi is θ=0,θ=2pi

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6(1-cos(θ))=sin^2(θ),0<= θ<= 2pi

Solution

6 (1 - cos (θ)) = sin^{2} (θ), 0 \leq θ \leq 2 π

Solution

θ = 0, θ = 2 π

Degrees

θ = 0^{\circ}, θ = 36 0^{\circ}

Solution steps

6 (1 - cos (θ)) = sin^{2} (θ), 0 \leq θ \leq 2 π

Subtract

sin^{2} (θ)

from both sides

6 (1 - cos (θ)) - sin^{2} (θ) = 0

Rewrite using trig identities

- sin^{2} (θ) + (1 - cos (θ)) \cdot 6

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 6

Simplify

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 6 : cos^{2} (θ) - 6 cos (θ) + 5

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 6

= - (1 - cos^{2} (θ)) + 6 (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 6

Expand

6 (1 - cos (θ)) : 6 - 6 cos (θ)

6 (1 - cos (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 1, c = cos (θ)

= 6 \cdot 1 - 6 cos (θ)

Multiply the numbers:

6 \cdot 1 = 6

= 6 - 6 cos (θ)

= - 1 + cos^{2} (θ) + 6 - 6 cos (θ)

Simplify

- 1 + cos^{2} (θ) + 6 - 6 cos (θ) : cos^{2} (θ) - 6 cos (θ) + 5

- 1 + cos^{2} (θ) + 6 - 6 cos (θ)

Group like terms

= cos^{2} (θ) - 6 cos (θ) - 1 + 6

Add/Subtract the numbers:

- 1 + 6 = 5

= cos^{2} (θ) - 6 cos (θ) + 5

= cos^{2} (θ) - 6 cos (θ) + 5

= cos^{2} (θ) - 6 cos (θ) + 5

5 + cos^{2} (θ) - 6 cos (θ) = 0

Solve by substitution

5 + cos^{2} (θ) - 6 cos (θ) = 0

Let:

cos (θ) = u

5 + u^{2} - 6 u = 0

5 + u^{2} - 6 u = 0 : u = 5, u = 1

5 + u^{2} - 6 u = 0

Write in the standard form

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

u^{2} - 6 u + 5 = 0

Solve with the quadratic formula

u^{2} - 6 u + 5 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = 5

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 5 = 4

(- 6)^{2} - 4 \cdot 1 \cdot 5

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 5

Multiply the numbers:

4 \cdot 1 \cdot 5 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

Subtract the numbers:

36 - 20 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 6 ) \pm 4}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 4}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 4}{2 \cdot 1}

u = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

\frac{- ( - 6 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 4}{2 \cdot 1}

Add the numbers:

6 + 4 = 10

= \frac{10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

\frac{- ( - 6 ) - 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 4}{2 \cdot 1}

Subtract the numbers:

6 - 4 = 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 = 5, u = 1

Substitute back

u = cos (θ)

cos (θ) = 5, cos (θ) = 1

cos (θ) = 5, cos (θ) = 1

cos (θ) = 5, 0 \leq θ \leq 2 π :

No Solution

cos (θ) = 5, 0 \leq θ \leq 2 π

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = 1, 0 \leq θ \leq 2 π : θ = 0, θ = 2 π

cos (θ) = 1, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = 0, θ = 2 π

Combine all the solutions

θ = 0, θ = 2 π

Popular Examples

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5cos^2(x)+14sin(x)cos(x)=3sin^2(x)

5 cos^{2} (x) + 14 sin (x) cos (x) = 3 sin^{2} (x)

4cos^2(x)+4sin^2(x)-5=0

4 cos^{2} (x) + 4 sin^{2} (x) - 5 = 0

4sin^2(x)-7cos(x)=2

4 sin^{2} (x) - 7 cos (x) = 2

Frequently Asked Questions (FAQ)

What is the general solution for 6(1-cos(θ))=sin^2(θ),0<= θ<= 2pi ?
The general solution for 6(1-cos(θ))=sin^2(θ),0<= θ<= 2pi is θ=0,θ=2pi