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

The general solution for 4cos(θ/2)+3=-2cos(θ) is θ=(4pi)/3+4pin,θ=(8pi)/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{4 π}{3} + 4 πn, θ = \frac{8 π}{3} + 4 πn

Degrees

θ = 24 0^{\circ} + 72 0^{\circ} n, θ = 48 0^{\circ} + 72 0^{\circ} n

Solution steps

4 cos (\frac{θ}{2}) + 3 = - 2 cos (θ)

Subtract

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

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

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

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

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

4^{2} - 4 \cdot 4 \cdot 1 = 0

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 \cdot 4}

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

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

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

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

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

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

\frac{θ}{2} = \frac{2 π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{2 π}{3} + 2 πn

Multiply both sides by

2

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

Simplify

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

2 \cdot \frac{2 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

Multiply fractions:

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

= \frac{2 π 2}{3}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{4 π}{3} + 4 πn

θ = \frac{4 π}{3} + 4 πn

θ = \frac{4 π}{3} + 4 πn

θ = \frac{4 π}{3} + 4 πn

\frac{θ}{2} = \frac{4 π}{3} + 2 πn : θ = \frac{8 π}{3} + 4 πn

\frac{θ}{2} = \frac{4 π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{4 π}{3} + 2 πn

Multiply both sides by

2

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

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{4 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{4 π}{3} + 2 \cdot 2 πn : \frac{8 π}{3} + 4 πn

2 \cdot \frac{4 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{4 π}{3} = \frac{8 π}{3}

2 \cdot \frac{4 π}{3}

Multiply fractions:

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

= \frac{4 π 2}{3}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{8 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{8 π}{3} + 4 πn

θ = \frac{8 π}{3} + 4 πn

θ = \frac{8 π}{3} + 4 πn

θ = \frac{8 π}{3} + 4 πn

θ = \frac{4 π}{3} + 4 πn, θ = \frac{8 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=(4pi)/3+4pin,θ=(8pi)/3+4pin