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

The general solution for 3(1-cos(θ))=sin^2(θ) is θ=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3(1-cos(θ))=sin^2(θ)

Solution

3 (1 - cos (θ)) = sin^{2} (θ)

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

3 (1 - cos (θ)) = sin^{2} (θ)

Subtract

sin^{2} (θ)

from both sides

3 (1 - cos (θ)) - sin^{2} (θ) = 0

Rewrite using trig identities

- sin^{2} (θ) + (1 - cos (θ)) \cdot 3

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 3

Simplify

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

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

= - (1 - cos^{2} (θ)) + 3 (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 3

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)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 cos (θ)

= - 1 + cos^{2} (θ) + 3 - 3 cos (θ)

Simplify

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

- 1 + cos^{2} (θ) + 3 - 3 cos (θ)

Group like terms

= cos^{2} (θ) - 3 cos (θ) - 1 + 3

Add/Subtract the numbers:

- 1 + 3 = 2

= cos^{2} (θ) - 3 cos (θ) + 2

= cos^{2} (θ) - 3 cos (θ) + 2

= cos^{2} (θ) - 3 cos (θ) + 2

2 + cos^{2} (θ) - 3 cos (θ) = 0

Solve by substitution

2 + cos^{2} (θ) - 3 cos (θ) = 0

Let:

cos (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 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 \cdot 1}

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 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 = 2, u = 1

Substitute back

u = cos (θ)

cos (θ) = 2, cos (θ) = 1

cos (θ) = 2, cos (θ) = 1

cos (θ) = 2 :

No Solution

cos (θ) = 2

- 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

Popular Examples

3tan(x)-3=0

3 tan (x) - 3 = 0

2sin(θ/3)+sqrt(3)=0

2 sin (\frac{θ}{3}) + 3 = 0

6sin(4x)+4=5

6 sin (4 x) + 4 = 5

29cos(x)=-17-cos(x)

29 cos (x) = - 17 - cos (x)

2cos(x)+2cos(2x)=0

2 cos (x) + 2 cos (2 x) = 0

Frequently Asked Questions (FAQ)

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