What is the general solution for 3cos^2(θ)+6cos(θ)-4=0 ?

The general solution for 3cos^2(θ)+6cos(θ)-4=0 is θ=1.01511…+2pin,θ=2pi-1.01511…+2pin

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Popular >

3cos^2(θ)+6cos(θ)-4=0

Solution

3 cos^{2} (θ) + 6 cos (θ) - 4 = 0

Solution

θ = 1.01511 \dots + 2 πn, θ = 2 π - 1.01511 \dots + 2 πn

Degrees

θ = 58.16160 \dots^{\circ} + 36 0^{\circ} n, θ = 301.83839 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (θ) + 6 cos (θ) - 4 = 0

Solve by substitution

3 cos^{2} (θ) + 6 cos (θ) - 4 = 0

Let:

cos (θ) = u

3 u^{2} + 6 u - 4 = 0

3 u^{2} + 6 u - 4 = 0 : u = \frac{- 3 + 21}{3}, u = - \frac{3 + 21}{3}

3 u^{2} + 6 u - 4 = 0

Solve with the quadratic formula

3 u^{2} + 6 u - 4 = 0

Quadratic Equation Formula:

For

a = 3, b = 6, c = - 4

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

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

6^{2} - 4 \cdot 3 (- 4) = 221

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 6^{2} + 48

6^{2} = 36

= 36 + 48

Add the numbers:

36 + 48 = 84

= 84

Prime factorization of

84 : 2^{2} \cdot 3 \cdot 7

84

84

divides by

284 = 42 \cdot 2

= 2 \cdot 42

42

divides by

242 = 21 \cdot 2

= 2 \cdot 2 \cdot 21

21

divides by

321 = 7 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 7

2, 3, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 7

= 2^{2} \cdot 3 \cdot 7

= 2^{2} \cdot 3 \cdot 7

Apply radical rule:

= 2^{2} 3 \cdot 7

Apply radical rule:

2^{2} = 2

= 2 3 \cdot 7

Refine

= 221

u_{1, 2} = \frac{- 6 \pm 2 21}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 6 + 2 21}{2 \cdot 3}, u_{2} = \frac{- 6 - 2 21}{2 \cdot 3}

u = \frac{- 6 + 2 21}{2 \cdot 3} : \frac{- 3 + 21}{3}

\frac{- 6 + 2 21}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6 + 2 21}{6}

Factor

- 6 + 221 : 2 (- 3 + 21)

- 6 + 221

Rewrite as

= - 2 \cdot 3 + 221

Factor out common term

2

= 2 (- 3 + 21)

= \frac{2 ( - 3 + 21 )}{6}

Cancel the common factor:

2

= \frac{- 3 + 21}{3}

u = \frac{- 6 - 2 21}{2 \cdot 3} : - \frac{3 + 21}{3}

\frac{- 6 - 2 21}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6 - 2 21}{6}

Factor

- 6 - 221 : - 2 (3 + 21)

- 6 - 221

Rewrite as

= - 2 \cdot 3 - 221

Factor out common term

2

= - 2 (3 + 21)

= - \frac{2 ( 3 + 21 )}{6}

Cancel the common factor:

2

= - \frac{3 + 21}{3}

The solutions to the quadratic equation are:

u = \frac{- 3 + 21}{3}, u = - \frac{3 + 21}{3}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 3 + 21}{3}, cos (θ) = - \frac{3 + 21}{3}

cos (θ) = \frac{- 3 + 21}{3}, cos (θ) = - \frac{3 + 21}{3}

cos (θ) = \frac{- 3 + 21}{3} : θ = arccos (\frac{- 3 + 21}{3}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 21}{3}) + 2 πn

cos (θ) = \frac{- 3 + 21}{3}

Apply trig inverse properties

cos (θ) = \frac{- 3 + 21}{3}

General solutions for

cos (θ) = \frac{- 3 + 21}{3}

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

θ = arccos (\frac{- 3 + 21}{3}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 21}{3}) + 2 πn

θ = arccos (\frac{- 3 + 21}{3}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 21}{3}) + 2 πn

cos (θ) = - \frac{3 + 21}{3} :

No Solution

cos (θ) = - \frac{3 + 21}{3}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 3 + 21}{3}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 21}{3}) + 2 πn

Show solutions in decimal form

θ = 1.01511 \dots + 2 πn, θ = 2 π - 1.01511 \dots + 2 πn

Popular Examples

arccos(x)=1

arccos (x) = 1

cos(x)=sqrt(3/4)

cos (x) = \frac{3}{4}

sin(x)=sin(x/2)

sin (x) = sin (\frac{x}{2})

sec(x)-2=-tan(x)-3

sec (x) - 2 = - tan (x) - 3

-9.81+5(32.17)(12/2)sin(θ)=0

- 9.81 + 5 (32.17) (\frac{12}{2}) sin (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(θ)+6cos(θ)-4=0 ?
The general solution for 3cos^2(θ)+6cos(θ)-4=0 is θ=1.01511…+2pin,θ=2pi-1.01511…+2pin