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Popular >

-3cos^2(θ)-6cos(θ)+3=-cos(θ)+5

Solution

- 3 cos^{2} (θ) - 6 cos (θ) + 3 = - cos (θ) + 5

Solution

θ = π + 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 131.81031 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 3 cos^{2} (θ) - 6 cos (θ) + 3 = - cos (θ) + 5

Solve by substitution

- 3 cos^{2} (θ) - 6 cos (θ) + 3 = - cos (θ) + 5

Let:

cos (θ) = u

- 3 u^{2} - 6 u + 3 = - u + 5

- 3 u^{2} - 6 u + 3 = - u + 5 : u = - 1, u = - \frac{2}{3}

- 3 u^{2} - 6 u + 3 = - u + 5

Move

5

to the left side

- 3 u^{2} - 6 u + 3 = - u + 5

Subtract

5

from both sides

- 3 u^{2} - 6 u + 3 - 5 = - u + 5 - 5

Simplify

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

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

Move

u

to the left side

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

Add

u

to both sides

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

Simplify

- 3 u^{2} - 5 u - 2 = 0

- 3 u^{2} - 5 u - 2 = 0

Solve with the quadratic formula

- 3 u^{2} - 5 u - 2 = 0

Quadratic Equation Formula:

For

a = - 3, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 3 ) ( - 2 )}{2 ( - 3 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 3 ) ( - 2 )}{2 ( - 3 )}

(- 5)^{2} - 4 (- 3) (- 2) = 1

(- 5)^{2} - 4 (- 3) (- 2)

Apply rule

- (- a) = a

= (- 5)^{2} - 4 \cdot 3 \cdot 2

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 ( - 3 )}, u_{2} = \frac{- ( - 5 ) - 1}{2 ( - 3 )}

u = \frac{- ( - 5 ) + 1}{2 ( - 3 )} : - 1

\frac{- ( - 5 ) + 1}{2 ( - 3 )}

Remove parentheses:

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

= \frac{5 + 1}{- 2 \cdot 3}

Add the numbers:

5 + 1 = 6

= \frac{6}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{- 6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 5 ) - 1}{2 ( - 3 )} : - \frac{2}{3}

\frac{- ( - 5 ) - 1}{2 ( - 3 )}

Remove parentheses:

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

= \frac{5 - 1}{- 2 \cdot 3}

Subtract the numbers:

5 - 1 = 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 = - 1, u = - \frac{2}{3}

Substitute back

u = cos (θ)

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

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

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = π + 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

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