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

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

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Popular >

3cos^2(x)+cos(x)-4=0

Solution

3 cos^{2} (x) + cos (x) - 4 = 0

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (x) + cos (x) - 4 = 0

Solve by substitution

3 cos^{2} (x) + cos (x) - 4 = 0

Let:

cos (x) = u

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

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

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

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 4)

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 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} : 1

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

Add/Subtract the numbers:

- 1 + 7 = 6

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Subtract the numbers:

- 1 - 7 = - 8

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 8}{6}

Apply the fraction rule:

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

= - \frac{8}{6}

Cancel the common factor:

2

= - \frac{4}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - \frac{4}{3}

cos (x) = 1, cos (x) = - \frac{4}{3}

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

Popular Examples

sin(θ)=7

sin (θ) = 7

sin(3x-pi/2)=-(sqrt(3))/2 ,(3pi,4pi)

sin (3 x - \frac{π}{2}) = - \frac{3}{2}, (3 π, 4 π)

2sin^2(x)-7cos(x)-5=0

2 sin^{2} (x) - 7 cos (x) - 5 = 0

cos(x)=sin(0)

cos (x) = sin (0)

1= 1/(0.5)sin(x+30)

1 = \frac{1}{0.5} sin (x + 3 0^{\circ})

Frequently Asked Questions (FAQ)

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