What is the general solution for -3cos^2(θ)-10cos(θ)=-3cos(θ)-6 ?

The general solution for -3cos^2(θ)-10cos(θ)=-3cos(θ)-6 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin

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Popular >

-3cos^2(θ)-10cos(θ)=-3cos(θ)-6

Solution

- 3 cos^{2} (θ) - 10 cos (θ) = - 3 cos (θ) - 6

Solution

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

Degrees

θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 3 cos^{2} (θ) - 10 cos (θ) = - 3 cos (θ) - 6

Solve by substitution

- 3 cos^{2} (θ) - 10 cos (θ) = - 3 cos (θ) - 6

Let:

cos (θ) = u

- 3 u^{2} - 10 u = - 3 u - 6

- 3 u^{2} - 10 u = - 3 u - 6 : u = - 3, u = \frac{2}{3}

- 3 u^{2} - 10 u = - 3 u - 6

Move

6

to the left side

- 3 u^{2} - 10 u = - 3 u - 6

Add

6

to both sides

- 3 u^{2} - 10 u + 6 = - 3 u - 6 + 6

Simplify

- 3 u^{2} - 10 u + 6 = - 3 u

- 3 u^{2} - 10 u + 6 = - 3 u

Move

3 u

to the left side

- 3 u^{2} - 10 u + 6 = - 3 u

Add

3 u

to both sides

- 3 u^{2} - 10 u + 6 + 3 u = - 3 u + 3 u

Simplify

- 3 u^{2} - 7 u + 6 = 0

- 3 u^{2} - 7 u + 6 = 0

Solve with the quadratic formula

- 3 u^{2} - 7 u + 6 = 0

Quadratic Equation Formula:

For

a = - 3, b = - 7, c = 6

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 3 ) \cdot 6}{2 ( - 3 )}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 3 ) \cdot 6}{2 ( - 3 )}

(- 7)^{2} - 4 (- 3) \cdot 6 = 11

(- 7)^{2} - 4 (- 3) \cdot 6

Apply rule

- (- a) = a

= (- 7)^{2} + 4 \cdot 3 \cdot 6

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 3 \cdot 6

Multiply the numbers:

4 \cdot 3 \cdot 6 = 72

= 7^{2} + 72

7^{2} = 49

= 49 + 72

Add the numbers:

49 + 72 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 7 ) \pm 11}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 11}{2 ( - 3 )}, u_{2} = \frac{- ( - 7 ) - 11}{2 ( - 3 )}

u = \frac{- ( - 7 ) + 11}{2 ( - 3 )} : - 3

\frac{- ( - 7 ) + 11}{2 ( - 3 )}

Remove parentheses:

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

= \frac{7 + 11}{- 2 \cdot 3}

Add the numbers:

7 + 11 = 18

= \frac{18}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{18}{- 6}

Apply the fraction rule:

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

= - \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= - 3

u = \frac{- ( - 7 ) - 11}{2 ( - 3 )} : \frac{2}{3}

\frac{- ( - 7 ) - 11}{2 ( - 3 )}

Remove parentheses:

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

= \frac{7 - 11}{- 2 \cdot 3}

Subtract the numbers:

7 - 11 = - 4

= \frac{- 4}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 4}{- 6}

Apply the fraction rule:

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

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = - 3, cos (θ) = \frac{2}{3}

cos (θ) = - 3, cos (θ) = \frac{2}{3}

cos (θ) = - 3 :

No Solution

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

No Solution

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

cos (θ) = \frac{2}{3}

Apply trig inverse properties

cos (θ) = \frac{2}{3}

General solutions for

cos (θ) = \frac{2}{3}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -3cos^2(θ)-10cos(θ)=-3cos(θ)-6 ?
The general solution for -3cos^2(θ)-10cos(θ)=-3cos(θ)-6 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin