What is the general solution for 7cos^2(θ)-3=-2cos(θ)+2 ?

The general solution for 7cos^2(θ)-3=-2cos(θ)+2 is θ=0.77519…+2pin,θ=2pi-0.77519…+2pin,θ=pi+2pin

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7cos^2(θ)-3=-2cos(θ)+2

Solution

7 cos^{2} (θ) - 3 = - 2 cos (θ) + 2

Solution

θ = 0.77519 \dots + 2 πn, θ = 2 π - 0.77519 \dots + 2 πn, θ = π + 2 πn

Degrees

θ = 44.41530 \dots^{\circ} + 36 0^{\circ} n, θ = 315.58469 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

7 cos^{2} (θ) - 3 = - 2 cos (θ) + 2

Solve by substitution

7 cos^{2} (θ) - 3 = - 2 cos (θ) + 2

Let:

cos (θ) = u

7 u^{2} - 3 = - 2 u + 2

7 u^{2} - 3 = - 2 u + 2 : u = \frac{5}{7}, u = - 1

7 u^{2} - 3 = - 2 u + 2

Move

2

to the left side

7 u^{2} - 3 = - 2 u + 2

Subtract

2

from both sides

7 u^{2} - 3 - 2 = - 2 u + 2 - 2

Simplify

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

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

Move

2 u

to the left side

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

Add

2 u

to both sides

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

Simplify

7 u^{2} - 5 + 2 u = 0

7 u^{2} - 5 + 2 u = 0

Write in the standard form

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

7 u^{2} + 2 u - 5 = 0

Solve with the quadratic formula

7 u^{2} + 2 u - 5 = 0

Quadratic Equation Formula:

For

a = 7, b = 2, c = - 5

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 7 ( - 5 )}{2 \cdot 7}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 7 ( - 5 )}{2 \cdot 7}

2^{2} - 4 \cdot 7 (- 5) = 12

2^{2} - 4 \cdot 7 (- 5)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 7 \cdot 5

Multiply the numbers:

4 \cdot 7 \cdot 5 = 140

= 2^{2} + 140

2^{2} = 4

= 4 + 140

Add the numbers:

4 + 140 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- 2 \pm 12}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- 2 + 12}{2 \cdot 7}, u_{2} = \frac{- 2 - 12}{2 \cdot 7}

u = \frac{- 2 + 12}{2 \cdot 7} : \frac{5}{7}

\frac{- 2 + 12}{2 \cdot 7}

Add/Subtract the numbers:

- 2 + 12 = 10

= \frac{10}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{10}{14}

Cancel the common factor:

2

= \frac{5}{7}

u = \frac{- 2 - 12}{2 \cdot 7} : - 1

\frac{- 2 - 12}{2 \cdot 7}

Subtract the numbers:

- 2 - 12 = - 14

= \frac{- 14}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 14}{14}

Apply the fraction rule:

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

= - \frac{14}{14}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{5}{7}, u = - 1

Substitute back

u = cos (θ)

cos (θ) = \frac{5}{7}, cos (θ) = - 1

cos (θ) = \frac{5}{7}, cos (θ) = - 1

cos (θ) = \frac{5}{7} : θ = arccos (\frac{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 2 πn

cos (θ) = \frac{5}{7}

Apply trig inverse properties

cos (θ) = \frac{5}{7}

General solutions for

cos (θ) = \frac{5}{7}

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

θ = arccos (\frac{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 2 πn

θ = arccos (\frac{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 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{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 2 πn, θ = π + 2 πn

Show solutions in decimal form

θ = 0.77519 \dots + 2 πn, θ = 2 π - 0.77519 \dots + 2 πn, θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7cos^2(θ)-3=-2cos(θ)+2 ?
The general solution for 7cos^2(θ)-3=-2cos(θ)+2 is θ=0.77519…+2pin,θ=2pi-0.77519…+2pin,θ=pi+2pin