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

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

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Popular >

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

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

a x^{2} + b x + c = 0

- 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 ( - 2 ) \cdot 2}{2 ( - 2 )}

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

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

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

5^{2} = 5

= 5

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

Separate the solutions

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

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

\frac{- 3 + 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \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}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 3 - 5}{2 ( - 2 )} : 2

\frac{- 3 - 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \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} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

cot(x)= 5/12 ,pi<x<(3pi)/2

cot (x) = \frac{5}{12}, π < x < \frac{3 π}{2}

tan(x)= 1/5 ,tan(x+pi/2)

tan (x) = \frac{1}{5}, tan (x + \frac{π}{2})

2cos(5x)=0

2 cos (5 x) = 0

sin(θ)=-5/4

sin (θ) = - \frac{5}{4}

2sin^2(x/2)+cos(x)=1+sin(x)

2 sin^{2} (\frac{x}{2}) + cos (x) = 1 + sin (x)

Frequently Asked Questions (FAQ)

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