What is the general solution for 9cos^2(θ)-12cos(θ)=-4 ?

The general solution for 9cos^2(θ)-12cos(θ)=-4 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9cos^2(θ)-12cos(θ)=-4

Solution

9 cos^{2} (θ) - 12 cos (θ) = - 4

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

9 cos^{2} (θ) - 12 cos (θ) = - 4

Solve by substitution

9 cos^{2} (θ) - 12 cos (θ) = - 4

Let:

cos (θ) = u

9 u^{2} - 12 u = - 4

9 u^{2} - 12 u = - 4 : u = \frac{2}{3}

9 u^{2} - 12 u = - 4

Move

4

to the left side

9 u^{2} - 12 u = - 4

Add

4

to both sides

9 u^{2} - 12 u + 4 = - 4 + 4

Simplify

9 u^{2} - 12 u + 4 = 0

9 u^{2} - 12 u + 4 = 0

Solve with the quadratic formula

9 u^{2} - 12 u + 4 = 0

Quadratic Equation Formula:

For

a = 9, b = - 12, c = 4

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

(- 12)^{2} - 4 \cdot 9 \cdot 4 = 0

(- 12)^{2} - 4 \cdot 9 \cdot 4

Apply exponent rule:

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

n

is even

(- 12)^{2} = 1 2^{2}

= 1 2^{2} - 4 \cdot 9 \cdot 4

Multiply the numbers:

4 \cdot 9 \cdot 4 = 144

= 1 2^{2} - 144

1 2^{2} = 144

= 144 - 144

Subtract the numbers:

144 - 144 = 0

= 0

u_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 9}

u = \frac{- ( - 12 )}{2 \cdot 9}

\frac{- ( - 12 )}{2 \cdot 9} = \frac{2}{3}

\frac{- ( - 12 )}{2 \cdot 9}

Apply rule

- (- a) = a

= \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{2}{3}

The solution to the quadratic equation is:

u = \frac{2}{3}

Substitute back

u = cos (θ)

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{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

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

Popular Examples

-3sec(θ/3)=-3sqrt(2)

- 3 sec (\frac{θ}{3}) = - 32

4cos(2x-pi)-1=1

4 cos (2 x - π) - 1 = 1

csc(x)=4sin(x)

csc (x) = 4 sin (x)

tan(3+x)=0

tan (3 + x) = 0

2cot^2(x)-9csc(x)=3

2 cot^{2} (x) - 9 csc (x) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 9cos^2(θ)-12cos(θ)=-4 ?
The general solution for 9cos^2(θ)-12cos(θ)=-4 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin