What is the general solution for 2cos(2θ)=6cos(θ)-4 ?

The general solution for 2cos(2θ)=6cos(θ)-4 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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2cos(2θ)=6cos(θ)-4

Solution

2 cos (2 θ) = 6 cos (θ) - 4

Solution

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

Degrees

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

Solution steps

2 cos (2 θ) = 6 cos (θ) - 4

Subtract

6 cos (θ) - 4

from both sides

2 cos (2 θ) - 6 cos (θ) + 4 = 0

Rewrite using trig identities

4 + 2 cos (2 θ) - 6 cos (θ)

Use the Double Angle identity:

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

= 4 + 2 (2 cos^{2} (θ) - 1) - 6 cos (θ)

Simplify

4 + 2 (2 cos^{2} (θ) - 1) - 6 cos (θ) : 4 cos^{2} (θ) - 6 cos (θ) + 2

4 + 2 (2 cos^{2} (θ) - 1) - 6 cos (θ)

Expand

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

2 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (θ), c = 1

= 2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

Simplify

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

2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 2 = 4

= 4 cos^{2} (θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 4 cos^{2} (θ) - 2

= 4 cos^{2} (θ) - 2

= 4 + 4 cos^{2} (θ) - 2 - 6 cos (θ)

Simplify

4 + 4 cos^{2} (θ) - 2 - 6 cos (θ) : 4 cos^{2} (θ) - 6 cos (θ) + 2

4 + 4 cos^{2} (θ) - 2 - 6 cos (θ)

Group like terms

= 4 cos^{2} (θ) - 6 cos (θ) + 4 - 2

Add/Subtract the numbers:

4 - 2 = 2

= 4 cos^{2} (θ) - 6 cos (θ) + 2

= 4 cos^{2} (θ) - 6 cos (θ) + 2

= 4 cos^{2} (θ) - 6 cos (θ) + 2

2 + 4 cos^{2} (θ) - 6 cos (θ) = 0

Solve by substitution

2 + 4 cos^{2} (θ) - 6 cos (θ) = 0

Let:

cos (θ) = u

2 + 4 u^{2} - 6 u = 0

2 + 4 u^{2} - 6 u = 0 : u = 1, u = \frac{1}{2}

2 + 4 u^{2} - 6 u = 0

Write in the standard form

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

4 u^{2} - 6 u + 2 = 0

Solve with the quadratic formula

4 u^{2} - 6 u + 2 = 0

Quadratic Equation Formula:

For

a = 4, b = - 6, c = 2

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

(- 6)^{2} - 4 \cdot 4 \cdot 2 = 2

(- 6)^{2} - 4 \cdot 4 \cdot 2

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 2}{2 \cdot 4}, u_{2} = \frac{- ( - 6 ) - 2}{2 \cdot 4}

u = \frac{- ( - 6 ) + 2}{2 \cdot 4} : 1

\frac{- ( - 6 ) + 2}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{6 + 2}{2 \cdot 4}

Add the numbers:

6 + 2 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 6 ) - 2}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 6 ) - 2}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{6 - 2}{2 \cdot 4}

Subtract the numbers:

6 - 2 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1, cos (θ) = \frac{1}{2}

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ)=6cos(θ)-4 ?
The general solution for 2cos(2θ)=6cos(θ)-4 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin