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

Q: What is the general solution for 3cos^2(x)+cos(x)-2=0,0<= x<= 2pi ?

The general solution for 3cos^2(x)+cos(x)-2=0,0<= x<= 2pi is x=0.84106…,x=2pi-0.84106…,x=pi

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3cos^2(x)+cos(x)-2=0,0<= x<= 2pi

Solution

3 cos^{2} (x) + cos (x) - 2 = 0, 0 \leq x \leq 2 π

Solution

x = 0.84106 \dots, x = 2 π - 0.84106 \dots, x = π

Degrees

x = 48.18968 \dots^{\circ}, x = 311.81031 \dots^{\circ}, x = 18 0^{\circ}

Solution steps

3 cos^{2} (x) + cos (x) - 2 = 0, 0 \leq x \leq 2 π

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 3 (- 2) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 3}, u_{2} = \frac{- 1 - 5}{2 \cdot 3}

u = \frac{- 1 + 5}{2 \cdot 3} : \frac{2}{3}

\frac{- 1 + 5}{2 \cdot 3}

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

u = \frac{- 1 - 5}{2 \cdot 3} : - 1

\frac{- 1 - 5}{2 \cdot 3}

Subtract the numbers:

- 1 - 5 = - 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{2}{3}, u = - 1

Substitute back

u = cos (x)

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

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

cos (x) = \frac{2}{3}, 0 \leq x \leq 2 π : x = arccos (\frac{2}{3}), x = 2 π - arccos (\frac{2}{3})

cos (x) = \frac{2}{3}, 0 \leq x \leq 2 π

Apply trig inverse properties

cos (x) = \frac{2}{3}

General solutions for

cos (x) = \frac{2}{3}

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

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

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

Solutions for the range

0 \leq x \leq 2 π

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

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

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 = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π

Combine all the solutions

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

Show solutions in decimal form

x = 0.84106 \dots, x = 2 π - 0.84106 \dots, x = π

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)+cos(x)-2=0,0<= x<= 2pi ?
The general solution for 3cos^2(x)+cos(x)-2=0,0<= x<= 2pi is x=0.84106…,x=2pi-0.84106…,x=pi