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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3cos^2(θ)-cos(θ)-4=0

Solution

3 cos^{2} (θ) - cos (θ) - 4 = 0

Solution

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (θ) - cos (θ) - 4 = 0

Solve by substitution

3 cos^{2} (θ) - cos (θ) - 4 = 0

Let:

cos (θ) = u

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

3 u^{2} - u - 4 = 0 : u = \frac{4}{3}, u = - 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = - 1, c = - 4

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

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

(- 1)^{2} - 4 \cdot 3 (- 4) = 7

(- 1)^{2} - 4 \cdot 3 (- 4)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 3 \cdot 4

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 3 \cdot 4 = 48

4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 7}{2 \cdot 3} : \frac{4}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{1 + 7}{2 \cdot 3}

Add the numbers:

1 + 7 = 8

= \frac{8}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{8}{6}

Cancel the common factor:

2

= \frac{4}{3}

u = \frac{- ( - 1 ) - 7}{2 \cdot 3} : - 1

\frac{- ( - 1 ) - 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{1 - 7}{2 \cdot 3}

Subtract the numbers:

1 - 7 = - 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

The solutions to the quadratic equation are:

u = \frac{4}{3}, u = - 1

Substitute back

u = cos (θ)

cos (θ) = \frac{4}{3}, cos (θ) = - 1

cos (θ) = \frac{4}{3}, cos (θ) = - 1

cos (θ) = \frac{4}{3} :

No Solution

cos (θ) = \frac{4}{3}

- 1 \leq cos (x) \leq 1

No Solution

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

Combine all the solutions

θ = π + 2 πn

Popular Examples

sin(x)=(120*sin(8))/(30)

sin (x) = \frac{120 \cdot sin ( 8 )}{30}

sin(θ)=(sqrt(5))

sin (θ) = (5)

cos^2(x)-3cos(x)-2=0

cos^{2} (x) - 3 cos (x) - 2 = 0

tan(4x)=tan(2x)

tan (4 x) = tan (2 x)

solvefor x,cos(2x+5x)=cos(Hx)solve for

x, cos (2 x + 5 x) = cos (H x)

Frequently Asked Questions (FAQ)

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