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Popular >

4cos(2θ)+1=2cos(θ)

Solution

4 cos (2 θ) + 1 = 2 cos (θ)

Solution

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 41.40962 \dots^{\circ} + 36 0^{\circ} n, θ = 318.59037 \dots^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (2 θ) + 1 = 2 cos (θ)

Subtract

2 cos (θ)

from both sides

4 cos (2 θ) + 1 - 2 cos (θ) = 0

Rewrite using trig identities

1 - 2 cos (θ) + 4 cos (2 θ)

Use the Double Angle identity:

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

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

Simplify

1 - 2 cos (θ) + 4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 2 cos (θ) - 3

1 - 2 cos (θ) + 4 (2 cos^{2} (θ) - 1)

Expand

4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 4

4 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (θ), c = 1

= 4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

Simplify

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1 : 8 cos^{2} (θ) - 4

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (θ) - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 8 cos^{2} (θ) - 4

= 8 cos^{2} (θ) - 4

= 1 - 2 cos (θ) + 8 cos^{2} (θ) - 4

Simplify

1 - 2 cos (θ) + 8 cos^{2} (θ) - 4 : 8 cos^{2} (θ) - 2 cos (θ) - 3

1 - 2 cos (θ) + 8 cos^{2} (θ) - 4

Group like terms

= - 2 cos (θ) + 8 cos^{2} (θ) + 1 - 4

Add/Subtract the numbers:

1 - 4 = - 3

= 8 cos^{2} (θ) - 2 cos (θ) - 3

= 8 cos^{2} (θ) - 2 cos (θ) - 3

= 8 cos^{2} (θ) - 2 cos (θ) - 3

- 3 - 2 cos (θ) + 8 cos^{2} (θ) = 0

Solve by substitution

- 3 - 2 cos (θ) + 8 cos^{2} (θ) = 0

Let:

cos (θ) = u

- 3 - 2 u + 8 u^{2} = 0

- 3 - 2 u + 8 u^{2} = 0 : u = \frac{3}{4}, u = - \frac{1}{2}

- 3 - 2 u + 8 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 8, b = - 2, c = - 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 3) = 10

(- 2)^{2} - 4 \cdot 8 (- 3)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 8 \cdot 3

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 8 \cdot 3

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

Add the numbers:

4 + 96 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- ( - 2 ) \pm 10}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 10}{2 \cdot 8}, u_{2} = \frac{- ( - 2 ) - 10}{2 \cdot 8}

u = \frac{- ( - 2 ) + 10}{2 \cdot 8} : \frac{3}{4}

\frac{- ( - 2 ) + 10}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{2 + 10}{2 \cdot 8}

Add the numbers:

2 + 10 = 12

= \frac{12}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{12}{16}

Cancel the common factor:

4

= \frac{3}{4}

u = \frac{- ( - 2 ) - 10}{2 \cdot 8} : - \frac{1}{2}

\frac{- ( - 2 ) - 10}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{2 - 10}{2 \cdot 8}

Subtract the numbers:

2 - 10 = - 8

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 8}{16}

Apply the fraction rule:

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

= - \frac{8}{16}

Cancel the common factor:

8

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{3}{4}, u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{4}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{3}{4}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{3}{4} : θ = arccos (\frac{3}{4}) + 2 πn, θ = 2 π - arccos (\frac{3}{4}) + 2 πn

cos (θ) = \frac{3}{4}

Apply trig inverse properties

cos (θ) = \frac{3}{4}

General solutions for

cos (θ) = \frac{3}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{3}{4}) + 2 πn, θ = 2 π - arccos (\frac{3}{4}) + 2 πn

θ = arccos (\frac{3}{4}) + 2 πn, θ = 2 π - arccos (\frac{3}{4}) + 2 πn

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

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

Combine all the solutions

θ = arccos (\frac{3}{4}) + 2 πn, θ = 2 π - arccos (\frac{3}{4}) + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Show solutions in decimal form

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

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