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Popular >

9cos^2(θ)-10cos(θ)-2=-7cos(θ)

Solution

9 cos^{2} (θ) - 10 cos (θ) - 2 = - 7 cos (θ)

Solution

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

Degrees

θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 109.47122 \dots^{\circ} + 36 0^{\circ} n, θ = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

9 cos^{2} (θ) - 10 cos (θ) - 2 = - 7 cos (θ)

Solve by substitution

9 cos^{2} (θ) - 10 cos (θ) - 2 = - 7 cos (θ)

Let:

cos (θ) = u

9 u^{2} - 10 u - 2 = - 7 u

9 u^{2} - 10 u - 2 = - 7 u : u = \frac{2}{3}, u = - \frac{1}{3}

9 u^{2} - 10 u - 2 = - 7 u

Move

7 u

to the left side

9 u^{2} - 10 u - 2 = - 7 u

Add

7 u

to both sides

9 u^{2} - 10 u - 2 + 7 u = - 7 u + 7 u

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 9, b = - 3, c = - 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 2 )}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 2 )}{2 \cdot 9}

(- 3)^{2} - 4 \cdot 9 (- 2) = 9

(- 3)^{2} - 4 \cdot 9 (- 2)

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 9 \cdot 2

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 9 \cdot 2

Multiply the numbers:

4 \cdot 9 \cdot 2 = 72

= 3^{2} + 72

3^{2} = 9

= 9 + 72

Add the numbers:

9 + 72 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 9}, u_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 9}

u = \frac{- ( - 3 ) + 9}{2 \cdot 9} : \frac{2}{3}

\frac{- ( - 3 ) + 9}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{3 + 9}{2 \cdot 9}

Add the numbers:

3 + 9 = 12

= \frac{12}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{12}{18}

Cancel the common factor:

6

= \frac{2}{3}

u = \frac{- ( - 3 ) - 9}{2 \cdot 9} : - \frac{1}{3}

\frac{- ( - 3 ) - 9}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{3 - 9}{2 \cdot 9}

Subtract the numbers:

3 - 9 = - 6

= \frac{- 6}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 6}{18}

Apply the fraction rule:

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

= - \frac{6}{18}

Cancel the common factor:

6

= - \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

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

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

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

cos (θ) = - \frac{1}{3} : θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

cos (θ) = - \frac{1}{3}

Apply trig inverse properties

cos (θ) = - \frac{1}{3}

General solutions for

cos (θ) = - \frac{1}{3}

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

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

Combine all the solutions

θ = arccos (\frac{2}{3}) + 2 πn, θ = 2 π - arccos (\frac{2}{3}) + 2 πn, θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

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