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

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

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

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

Solution

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

Solution

x = \frac{π}{3}, x = - \frac{π}{3}

Degrees

x = 6 0^{\circ}, x = - 6 0^{\circ}

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 5 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Subtract the numbers:

- 3 - 5 = - 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 2

Substitute back

u = cos (x)

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

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

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

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

General solutions for

cos (x) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solutions for the range

- π \leq x \leq π

x = \frac{π}{3}, x = - \frac{π}{3}

cos (x) = - 2, - π \leq x \leq π :

No Solution

cos (x) = - 2, - π \leq x \leq π

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{3}, x = - \frac{π}{3}

Graph

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Frequently Asked Questions (FAQ)

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