What is the general solution for 4cos^2(θ)-cos(θ)-9=7cos(θ)-4 ?

The general solution for 4cos^2(θ)-cos(θ)-9=7cos(θ)-4 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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Popular >

4cos^2(θ)-cos(θ)-9=7cos(θ)-4

Solution

4 cos^{2} (θ) - cos (θ) - 9 = 7 cos (θ) - 4

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 = 7 cos (θ) - 4

Solve by substitution

4 cos^{2} (θ) - cos (θ) - 9 = 7 cos (θ) - 4

Let:

cos (θ) = u

4 u^{2} - u - 9 = 7 u - 4

4 u^{2} - u - 9 = 7 u - 4 : u = \frac{5}{2}, u = - \frac{1}{2}

4 u^{2} - u - 9 = 7 u - 4

Move

4

to the left side

4 u^{2} - u - 9 = 7 u - 4

Add

4

to both sides

4 u^{2} - u - 9 + 4 = 7 u - 4 + 4

Simplify

4 u^{2} - u - 5 = 7 u

4 u^{2} - u - 5 = 7 u

Move

7 u

to the left side

4 u^{2} - u - 5 = 7 u

Subtract

7 u

from both sides

4 u^{2} - u - 5 - 7 u = 7 u - 7 u

Simplify

4 u^{2} - 8 u - 5 = 0

4 u^{2} - 8 u - 5 = 0

Solve with the quadratic formula

4 u^{2} - 8 u - 5 = 0

Quadratic Equation Formula:

For

a = 4, b = - 8, c = - 5

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 (- 5) = 12

(- 8)^{2} - 4 \cdot 4 (- 5)

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 4 \cdot 5

Apply exponent rule:

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

n

is even

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

= 8^{2} + 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

Add the numbers:

64 + 80 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 8 ) \pm 12}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 12}{2 \cdot 4}, u_{2} = \frac{- ( - 8 ) - 12}{2 \cdot 4}

u = \frac{- ( - 8 ) + 12}{2 \cdot 4} : \frac{5}{2}

\frac{- ( - 8 ) + 12}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{8 + 12}{2 \cdot 4}

Add the numbers:

8 + 12 = 20

= \frac{20}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{20}{8}

Cancel the common factor:

4

= \frac{5}{2}

u = \frac{- ( - 8 ) - 12}{2 \cdot 4} : - \frac{1}{2}

\frac{- ( - 8 ) - 12}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{8 - 12}{2 \cdot 4}

Subtract the numbers:

8 - 12 = - 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{5}{2}, u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{5}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{5}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{5}{2} :

No Solution

cos (θ) = \frac{5}{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

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cos (x) = 0.998482

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cos(5x)-sin(2x)-cos(x)=0

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(θ)-cos(θ)-9=7cos(θ)-4 ?
The general solution for 4cos^2(θ)-cos(θ)-9=7cos(θ)-4 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin