What is the general solution for 4cos^2(θ)+3cos(θ)=1 ?

The general solution for 4cos^2(θ)+3cos(θ)=1 is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin,θ=pi+2pin

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Popular Trigonometry >

4cos^2(θ)+3cos(θ)=1

Solution

4 cos^{2} (θ) + 3 cos (θ) = 1

Solution

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn, θ = π + 2 πn

Degrees

θ = 75.52248 \dots^{\circ} + 36 0^{\circ} n, θ = 284.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (θ) + 3 cos (θ) = 1

Solve by substitution

4 cos^{2} (θ) + 3 cos (θ) = 1

Let:

cos (θ) = u

4 u^{2} + 3 u = 1

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

4 u^{2} + 3 u = 1

Move

1

to the left side

4 u^{2} + 3 u = 1

Subtract

1

from both sides

4 u^{2} + 3 u - 1 = 1 - 1

Simplify

4 u^{2} + 3 u - 1 = 0

4 u^{2} + 3 u - 1 = 0

Solve with the quadratic formula

4 u^{2} + 3 u - 1 = 0

Quadratic Equation Formula:

For

a = 4, b = 3, c = - 1

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

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

3^{2} - 4 \cdot 4 (- 1) = 5

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 3 + 5}{2 \cdot 4}, u_{2} = \frac{- 3 - 5}{2 \cdot 4}

u = \frac{- 3 + 5}{2 \cdot 4} : \frac{1}{4}

\frac{- 3 + 5}{2 \cdot 4}

Add/Subtract the numbers:

- 3 + 5 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

u = \frac{- 3 - 5}{2 \cdot 4} : - 1

\frac{- 3 - 5}{2 \cdot 4}

Subtract the numbers:

- 3 - 5 = - 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{4}, cos (θ) = - 1

cos (θ) = \frac{1}{4}, cos (θ) = - 1

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

cos (θ) = \frac{1}{4}

Apply trig inverse properties

cos (θ) = \frac{1}{4}

General solutions for

cos (θ) = \frac{1}{4}

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

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

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

cos (θ) = - 1 : θ = π + 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 = π + 2 πn

θ = π + 2 πn

Combine all the solutions

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

Show solutions in decimal form

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn, θ = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(θ)+3cos(θ)=1 ?
The general solution for 4cos^2(θ)+3cos(θ)=1 is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin,θ=pi+2pin