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

The general solution for 2*sin^2(θ)=3(1-cos(θ)) is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=2pin

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2*sin^2(θ)=3(1-cos(θ))

Solution

2 \cdot sin^{2} (θ) = 3 (1 - cos (θ))

Solution

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (θ) = 3 (1 - cos (θ))

Subtract

3 (1 - cos (θ))

from both sides

2 sin^{2} (θ) - 3 (1 - cos (θ)) = 0

Rewrite using trig identities

- (1 - cos (θ)) \cdot 3 + 2 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - (1 - cos (θ)) \cdot 3 + 2 (1 - cos^{2} (θ))

Simplify

- (1 - cos (θ)) \cdot 3 + 2 (1 - cos^{2} (θ)) : 3 cos (θ) - 2 cos^{2} (θ) - 1

- (1 - cos (θ)) \cdot 3 + 2 (1 - cos^{2} (θ))

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

Expand

- 3 (1 - cos (θ)) : - 3 + 3 cos (θ)

- 3 (1 - cos (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos (θ)

= - 3 \cdot 1 - (- 3) cos (θ)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 cos (θ)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 cos (θ)

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

Expand

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

2 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (θ)

= 2 \cdot 1 - 2 cos^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (θ)

= - 3 + 3 cos (θ) + 2 - 2 cos^{2} (θ)

Simplify

- 3 + 3 cos (θ) + 2 - 2 cos^{2} (θ) : 3 cos (θ) - 2 cos^{2} (θ) - 1

- 3 + 3 cos (θ) + 2 - 2 cos^{2} (θ)

Group like terms

= 3 cos (θ) - 2 cos^{2} (θ) - 3 + 2

Add/Subtract the numbers:

- 3 + 2 = - 1

= 3 cos (θ) - 2 cos^{2} (θ) - 1

= 3 cos (θ) - 2 cos^{2} (θ) - 1

= 3 cos (θ) - 2 cos^{2} (θ) - 1

- 1 - 2 cos^{2} (θ) + 3 cos (θ) = 0

Solve by substitution

- 1 - 2 cos^{2} (θ) + 3 cos (θ) = 0

Let:

cos (θ) = u

- 1 - 2 u^{2} + 3 u = 0

- 1 - 2 u^{2} + 3 u = 0 : u = \frac{1}{2}, u = 1

- 1 - 2 u^{2} + 3 u = 0

Write in the standard form

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

- 2 u^{2} + 3 u - 1 = 0

Solve with the quadratic formula

- 2 u^{2} + 3 u - 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 3, c = - 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

3^{2} - 4 (- 2) (- 1) = 1

3^{2} - 4 (- 2) (- 1)

Apply rule

- (- a) = a

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 ( - 2 )}, u_{2} = \frac{- 3 - 1}{2 ( - 2 )}

u = \frac{- 3 + 1}{2 ( - 2 )} : \frac{1}{2}

\frac{- 3 + 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 1}{- 2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 1}{2 ( - 2 )} : 1

\frac{- 3 - 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 1}{- 2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 1

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2}, cos (θ) = 1

cos (θ) = \frac{1}{2}, cos (θ) = 1

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 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 = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2*sin^2(θ)=3(1-cos(θ)) ?
The general solution for 2*sin^2(θ)=3(1-cos(θ)) is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=2pin