What is the general solution for 6cos^2(θ)+7=-17cos(θ) ?

The general solution for 6cos^2(θ)+7=-17cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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6cos^2(θ)+7=-17cos(θ)

Solution

6 cos^{2} (θ) + 7 = - 17 cos (θ)

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

6 cos^{2} (θ) + 7 = - 17 cos (θ)

Solve by substitution

6 cos^{2} (θ) + 7 = - 17 cos (θ)

Let:

cos (θ) = u

6 u^{2} + 7 = - 17 u

6 u^{2} + 7 = - 17 u : u = - \frac{1}{2}, u = - \frac{7}{3}

6 u^{2} + 7 = - 17 u

Move

17 u

to the left side

6 u^{2} + 7 = - 17 u

Add

17 u

to both sides

6 u^{2} + 7 + 17 u = - 17 u + 17 u

Simplify

6 u^{2} + 7 + 17 u = 0

6 u^{2} + 7 + 17 u = 0

Write in the standard form

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

6 u^{2} + 17 u + 7 = 0

Solve with the quadratic formula

6 u^{2} + 17 u + 7 = 0

Quadratic Equation Formula:

For

a = 6, b = 17, c = 7

u_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 6 \cdot 7}{2 \cdot 6}

u_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 6 \cdot 7}{2 \cdot 6}

1 7^{2} - 4 \cdot 6 \cdot 7 = 11

1 7^{2} - 4 \cdot 6 \cdot 7

Multiply the numbers:

4 \cdot 6 \cdot 7 = 168

= 1 7^{2} - 168

1 7^{2} = 289

= 289 - 168

Subtract the numbers:

289 - 168 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 17 \pm 11}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 17 + 11}{2 \cdot 6}, u_{2} = \frac{- 17 - 11}{2 \cdot 6}

u = \frac{- 17 + 11}{2 \cdot 6} : - \frac{1}{2}

\frac{- 17 + 11}{2 \cdot 6}

Add/Subtract the numbers:

- 17 + 11 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

u = \frac{- 17 - 11}{2 \cdot 6} : - \frac{7}{3}

\frac{- 17 - 11}{2 \cdot 6}

Subtract the numbers:

- 17 - 11 = - 28

= \frac{- 28}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 28}{12}

Apply the fraction rule:

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

= - \frac{28}{12}

Cancel the common factor:

4

= - \frac{7}{3}

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = - \frac{7}{3}

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}, cos (θ) = - \frac{7}{3}

cos (θ) = - \frac{1}{2}, cos (θ) = - \frac{7}{3}

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

cos (θ) = - \frac{7}{3} :

No Solution

cos (θ) = - \frac{7}{3}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

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7 tan (4 x + 199) = 8 sin (4 x + 199)

Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(θ)+7=-17cos(θ) ?
The general solution for 6cos^2(θ)+7=-17cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin