What is the general solution for 4cos(θ/2)=3+2cos(θ) ?

The general solution for 4cos(θ/2)=3+2cos(θ) is θ=(2pi)/3+4pin,θ=(10pi)/3+4pin

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4cos(θ/2)=3+2cos(θ)

Solution

4 cos (\frac{θ}{2}) = 3 + 2 cos (θ)

Solution

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

Degrees

θ = 12 0^{\circ} + 72 0^{\circ} n, θ = 60 0^{\circ} + 72 0^{\circ} n

Solution steps

4 cos (\frac{θ}{2}) = 3 + 2 cos (θ)

Subtract

3 + 2 cos (θ)

from both sides

4 cos (\frac{θ}{2}) - 3 - 2 cos (θ) = 0

Let:

u = \frac{θ}{2}

4 cos (u) - 3 - 2 cos (2 u) = 0

Rewrite using trig identities

- 3 - 2 cos (2 u) + 4 cos (u)

Use the Double Angle identity:

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

= - 3 - 2 (2 cos^{2} (u) - 1) + 4 cos (u)

Simplify

- 3 - 2 (2 cos^{2} (u) - 1) + 4 cos (u) : 4 cos (u) - 4 cos^{2} (u) - 1

- 3 - 2 (2 cos^{2} (u) - 1) + 4 cos (u)

Expand

- 2 (2 cos^{2} (u) - 1) : - 4 cos^{2} (u) + 2

- 2 (2 cos^{2} (u) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 2 cos^{2} (u), c = 1

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

Apply minus-plus rules

- (- a) = a

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

Simplify

- 2 \cdot 2 cos^{2} (u) + 2 \cdot 1 : - 4 cos^{2} (u) + 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= - 4 cos^{2} (u) + 2

= - 4 cos^{2} (u) + 2

= - 3 - 4 cos^{2} (u) + 2 + 4 cos (u)

Simplify

- 3 - 4 cos^{2} (u) + 2 + 4 cos (u) : 4 cos (u) - 4 cos^{2} (u) - 1

- 3 - 4 cos^{2} (u) + 2 + 4 cos (u)

Group like terms

= - 4 cos^{2} (u) + 4 cos (u) - 3 + 2

Add/Subtract the numbers:

- 3 + 2 = - 1

= 4 cos (u) - 4 cos^{2} (u) - 1

= 4 cos (u) - 4 cos^{2} (u) - 1

= 4 cos (u) - 4 cos^{2} (u) - 1

- 1 + 4 cos (u) - 4 cos^{2} (u) = 0

Solve by substitution

- 1 + 4 cos (u) - 4 cos^{2} (u) = 0

Let:

cos (u) = u

- 1 + 4 u - 4 u^{2} = 0

- 1 + 4 u - 4 u^{2} = 0 : u = \frac{1}{2}

- 1 + 4 u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 4, c = - 1

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

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

4^{2} - 4 (- 4) (- 1) = 0

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 ( - 4 )}

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

\frac{- 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = cos (u)

cos (u) = \frac{1}{2}

cos (u) = \frac{1}{2}

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

cos (u) = \frac{1}{2}

General solutions for

cos (u) = \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}

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

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

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

\frac{θ}{2} = \frac{π}{3} + 2 πn : θ = \frac{2 π}{3} + 4 πn

\frac{θ}{2} = \frac{π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{3} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{3} + 2 \cdot 2 πn : \frac{2 π}{3} + 4 πn

2 \cdot \frac{π}{3} + 2 \cdot 2 πn

Multiply

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{3} + 2 πn : θ = \frac{10 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn : \frac{10 π}{3} + 4 πn

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{3} = \frac{10 π}{3}

2 \cdot \frac{5 π}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 π 2}{3}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(θ/2)=3+2cos(θ) ?
The general solution for 4cos(θ/2)=3+2cos(θ) is θ=(2pi)/3+4pin,θ=(10pi)/3+4pin