What is the general solution for 4cos^2(θ)-5cos(θ)=-4cos(θ)+3 ?

The general solution for 4cos^2(θ)-5cos(θ)=-4cos(θ)+3 is θ=2pin,θ=2.41885…+2pin,θ=-2.41885…+2pin

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4cos^2(θ)-5cos(θ)=-4cos(θ)+3

Solution

4 cos^{2} (θ) - 5 cos (θ) = - 4 cos (θ) + 3

Solution

θ = 2 πn, θ = 2.41885 \dots + 2 πn, θ = - 2.41885 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 138.59037 \dots^{\circ} + 36 0^{\circ} n, θ = - 138.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (θ) - 5 cos (θ) = - 4 cos (θ) + 3

Solve by substitution

4 cos^{2} (θ) - 5 cos (θ) = - 4 cos (θ) + 3

Let:

cos (θ) = u

4 u^{2} - 5 u = - 4 u + 3

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

4 u^{2} - 5 u = - 4 u + 3

Move

3

to the left side

4 u^{2} - 5 u = - 4 u + 3

Subtract

3

from both sides

4 u^{2} - 5 u - 3 = - 4 u + 3 - 3

Simplify

4 u^{2} - 5 u - 3 = - 4 u

4 u^{2} - 5 u - 3 = - 4 u

Move

4 u

to the left side

4 u^{2} - 5 u - 3 = - 4 u

Add

4 u

to both sides

4 u^{2} - 5 u - 3 + 4 u = - 4 u + 4 u

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 \cdot 4 (- 3) = 7

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 4 \cdot 3 = 48

4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 4}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 4}

u = \frac{- ( - 1 ) + 7}{2 \cdot 4} : 1

\frac{- ( - 1 ) + 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 + 7}{2 \cdot 4}

Add the numbers:

1 + 7 = 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{- ( - 1 ) - 7}{2 \cdot 4} : - \frac{3}{4}

\frac{- ( - 1 ) - 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 - 7}{2 \cdot 4}

Subtract the numbers:

1 - 7 = - 6

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 6}{8}

Apply the fraction rule:

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

= - \frac{6}{8}

Cancel the common factor:

2

= - \frac{3}{4}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = - \frac{3}{4}

cos (θ) = 1, cos (θ) = - \frac{3}{4}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 2 πn, θ = 2.41885 \dots + 2 πn, θ = - 2.41885 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(θ)-5cos(θ)=-4cos(θ)+3 ?
The general solution for 4cos^2(θ)-5cos(θ)=-4cos(θ)+3 is θ=2pin,θ=2.41885…+2pin,θ=-2.41885…+2pin