What is the general solution for 2sin(θ)=3(1-cos(θ)) ?

The general solution for 2sin(θ)=3(1-cos(θ)) is θ=1.17600…+2pin,θ=2pin

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2sin(θ)=3(1-cos(θ))

Solution

2 sin (θ) = 3 (1 - cos (θ))

Solution

θ = 1.17600 \dots + 2 πn, θ = 2 πn

Degrees

θ = 67.38013 \dots^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) = 3 (1 - cos (θ))

Square both sides

(2 sin (θ))^{2} = (3 (1 - cos (θ)))^{2}

Subtract

(3 (1 - cos (θ)))^{2}

from both sides

4 sin^{2} (θ) - 9 (1 - cos (θ))^{2} = 0

Rewrite using trig identities

- (1 - cos (θ))^{2} \cdot 9 + 4 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - (1 - cos (θ))^{2} \cdot 9 + 4 (1 - cos^{2} (θ))

Simplify

- (1 - cos (θ))^{2} \cdot 9 + 4 (1 - cos^{2} (θ)) : 18 cos (θ) - 13 cos^{2} (θ) - 5

- (1 - cos (θ))^{2} \cdot 9 + 4 (1 - cos^{2} (θ))

= - 9 (1 - cos (θ))^{2} + 4 (1 - cos^{2} (θ))

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = cos (θ)

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

= 1 - 2 cos (θ) + cos^{2} (θ)

= 1 - 2 cos (θ) + cos^{2} (θ)

= - (1 - 2 cos (θ) + cos^{2} (θ)) \cdot 9 + 4 (1 - cos^{2} (θ))

Expand

- 9 (1 - 2 cos (θ) + cos^{2} (θ)) : - 9 + 18 cos (θ) - 9 cos^{2} (θ)

- 9 (1 - 2 cos (θ) + cos^{2} (θ))

Distribute parentheses

= (- 9) \cdot 1 + (- 9) (- 2 cos (θ)) + (- 9) cos^{2} (θ)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= - 9 \cdot 1 + 9 \cdot 2 cos (θ) - 9 cos^{2} (θ)

Simplify

- 9 \cdot 1 + 9 \cdot 2 cos (θ) - 9 cos^{2} (θ) : - 9 + 18 cos (θ) - 9 cos^{2} (θ)

- 9 \cdot 1 + 9 \cdot 2 cos (θ) - 9 cos^{2} (θ)

Multiply the numbers:

9 \cdot 1 = 9

= - 9 + 9 \cdot 2 cos (θ) - 9 cos^{2} (θ)

Multiply the numbers:

9 \cdot 2 = 18

= - 9 + 18 cos (θ) - 9 cos^{2} (θ)

= - 9 + 18 cos (θ) - 9 cos^{2} (θ)

= - 9 + 18 cos (θ) - 9 cos^{2} (θ) + 4 (1 - cos^{2} (θ))

Expand

4 (1 - cos^{2} (θ)) : 4 - 4 cos^{2} (θ)

4 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= - 9 + 18 cos (θ) - 9 cos^{2} (θ) + 4 - 4 cos^{2} (θ)

Simplify

- 9 + 18 cos (θ) - 9 cos^{2} (θ) + 4 - 4 cos^{2} (θ) : 18 cos (θ) - 13 cos^{2} (θ) - 5

- 9 + 18 cos (θ) - 9 cos^{2} (θ) + 4 - 4 cos^{2} (θ)

Group like terms

= 18 cos (θ) - 9 cos^{2} (θ) - 4 cos^{2} (θ) - 9 + 4

Add similar elements:

- 9 cos^{2} (θ) - 4 cos^{2} (θ) = - 13 cos^{2} (θ)

= 18 cos (θ) - 13 cos^{2} (θ) - 9 + 4

Add/Subtract the numbers:

- 9 + 4 = - 5

= 18 cos (θ) - 13 cos^{2} (θ) - 5

= 18 cos (θ) - 13 cos^{2} (θ) - 5

= 18 cos (θ) - 13 cos^{2} (θ) - 5

- 5 - 13 cos^{2} (θ) + 18 cos (θ) = 0

Solve by substitution

- 5 - 13 cos^{2} (θ) + 18 cos (θ) = 0

Let:

cos (θ) = u

- 5 - 13 u^{2} + 18 u = 0

- 5 - 13 u^{2} + 18 u = 0 : u = \frac{5}{13}, u = 1

- 5 - 13 u^{2} + 18 u = 0

Write in the standard form

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

- 13 u^{2} + 18 u - 5 = 0

Solve with the quadratic formula

- 13 u^{2} + 18 u - 5 = 0

Quadratic Equation Formula:

For

a = - 13, b = 18, c = - 5

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 13 ) ( - 5 )}{2 ( - 13 )}

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 13 ) ( - 5 )}{2 ( - 13 )}

1 8^{2} - 4 (- 13) (- 5) = 8

1 8^{2} - 4 (- 13) (- 5)

Apply rule

- (- a) = a

= 1 8^{2} - 4 \cdot 13 \cdot 5

Multiply the numbers:

4 \cdot 13 \cdot 5 = 260

= 1 8^{2} - 260

1 8^{2} = 324

= 324 - 260

Subtract the numbers:

324 - 260 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 18 \pm 8}{2 ( - 13 )}

Separate the solutions

u_{1} = \frac{- 18 + 8}{2 ( - 13 )}, u_{2} = \frac{- 18 - 8}{2 ( - 13 )}

u = \frac{- 18 + 8}{2 ( - 13 )} : \frac{5}{13}

\frac{- 18 + 8}{2 ( - 13 )}

Remove parentheses:

(- a) = - a

= \frac{- 18 + 8}{- 2 \cdot 13}

Add/Subtract the numbers:

- 18 + 8 = - 10

= \frac{- 10}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{- 10}{- 26}

Apply the fraction rule:

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

= \frac{10}{26}

Cancel the common factor:

2

= \frac{5}{13}

u = \frac{- 18 - 8}{2 ( - 13 )} : 1

\frac{- 18 - 8}{2 ( - 13 )}

Remove parentheses:

(- a) = - a

= \frac{- 18 - 8}{- 2 \cdot 13}

Subtract the numbers:

- 18 - 8 = - 26

= \frac{- 26}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{- 26}{- 26}

Apply the fraction rule:

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

= \frac{26}{26}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{5}{13}, u = 1

Substitute back

u = cos (θ)

cos (θ) = \frac{5}{13}, cos (θ) = 1

cos (θ) = \frac{5}{13}, cos (θ) = 1

cos (θ) = \frac{5}{13} : θ = arccos (\frac{5}{13}) + 2 πn, θ = 2 π - arccos (\frac{5}{13}) + 2 πn

cos (θ) = \frac{5}{13}

Apply trig inverse properties

cos (θ) = \frac{5}{13}

General solutions for

cos (θ) = \frac{5}{13}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{5}{13}) + 2 πn, θ = 2 π - arccos (\frac{5}{13}) + 2 πn

θ = arccos (\frac{5}{13}) + 2 πn, θ = 2 π - arccos (\frac{5}{13}) + 2 πn

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 = arccos (\frac{5}{13}) + 2 πn, θ = 2 π - arccos (\frac{5}{13}) + 2 πn, θ = 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 sin (θ) = 3 (1 - cos (θ))

Remove the ones that don't agree with the equation.

Check the solution

arccos (\frac{5}{13}) + 2 πn :

True

arccos (\frac{5}{13}) + 2 πn

Plug in

n = 1

arccos (\frac{5}{13}) + 2 π 1

For

2 sin (θ) = 3 (1 - cos (θ))

plug in

θ = arccos (\frac{5}{13}) + 2 π 1

2 sin (arccos (\frac{5}{13}) + 2 π 1) = 3 (1 - cos (arccos (\frac{5}{13}) + 2 π 1))

Refine

1.84615 \dots = 1.84615 \dots

\Rightarrow True

Check the solution

2 π - arccos (\frac{5}{13}) + 2 πn :

False

2 π - arccos (\frac{5}{13}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{5}{13}) + 2 π 1

For

2 sin (θ) = 3 (1 - cos (θ))

plug in

θ = 2 π - arccos (\frac{5}{13}) + 2 π 1

2 sin (2 π - arccos (\frac{5}{13}) + 2 π 1) = 3 (1 - cos (2 π - arccos (\frac{5}{13}) + 2 π 1))

Refine

- 1.84615 \dots = 1.84615 \dots

\Rightarrow False

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

2 sin (θ) = 3 (1 - cos (θ))

plug in

θ = 2 π 1

2 sin (2 π 1) = 3 (1 - cos (2 π 1))

Refine

0 = 0

\Rightarrow True

θ = arccos (\frac{5}{13}) + 2 πn, θ = 2 πn

Show solutions in decimal form

θ = 1.17600 \dots + 2 πn, θ = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(θ)=3(1-cos(θ)) ?
The general solution for 2sin(θ)=3(1-cos(θ)) is θ=1.17600…+2pin,θ=2pin