What is the general solution for -4cos^2(θ)+cos(θ)=9cos(θ)+3 ?

The general solution for -4cos^2(θ)+cos(θ)=9cos(θ)+3 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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Popular Trigonometry >

-4cos^2(θ)+cos(θ)=9cos(θ)+3

Solution

- 4 cos^{2} (θ) + cos (θ) = 9 cos (θ) + 3

Solution

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

Degrees

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

Solution steps

- 4 cos^{2} (θ) + cos (θ) = 9 cos (θ) + 3

Solve by substitution

- 4 cos^{2} (θ) + cos (θ) = 9 cos (θ) + 3

Let:

cos (θ) = u

- 4 u^{2} + u = 9 u + 3

- 4 u^{2} + u = 9 u + 3 : u = - \frac{3}{2}, u = - \frac{1}{2}

- 4 u^{2} + u = 9 u + 3

Move

3

to the left side

- 4 u^{2} + u = 9 u + 3

Subtract

3

from both sides

- 4 u^{2} + u - 3 = 9 u + 3 - 3

Simplify

- 4 u^{2} + u - 3 = 9 u

- 4 u^{2} + u - 3 = 9 u

Move

9 u

to the left side

- 4 u^{2} + u - 3 = 9 u

Subtract

9 u

from both sides

- 4 u^{2} + u - 3 - 9 u = 9 u - 9 u

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = - 8, c = - 3

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

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

(- 8)^{2} - 4 (- 4) (- 3) = 4

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

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

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

n a^{n} = a

4^{2} = 4

= 4

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

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 4}{2 ( - 4 )}, u_{2} = \frac{- ( - 8 ) - 4}{2 ( - 4 )}

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

\frac{- ( - 8 ) + 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add the numbers:

8 + 4 = 12

= \frac{12}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{12}{- 8}

Apply the fraction rule:

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

= - \frac{12}{8}

Cancel the common factor:

4

= - \frac{3}{2}

u = \frac{- ( - 8 ) - 4}{2 ( - 4 )} : - \frac{1}{2}

\frac{- ( - 8 ) - 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Subtract the numbers:

8 - 4 = 4

= \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}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

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

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

cos (θ) = - \frac{3}{2} :

No Solution

cos (θ) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -4cos^2(θ)+cos(θ)=9cos(θ)+3 ?
The general solution for -4cos^2(θ)+cos(θ)=9cos(θ)+3 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin