What is the general solution for -4cos^2(x)+6cos(x)=2 ?

The general solution for -4cos^2(x)+6cos(x)=2 is x= pi/3+2pin,x=(5pi)/3+2pin,x=2pin

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Popular Trigonometry >

-4cos^2(x)+6cos(x)=2

Solution

- 4 cos^{2} (x) + 6 cos (x) = 2

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

- 4 cos^{2} (x) + 6 cos (x) = 2

Solve by substitution

- 4 cos^{2} (x) + 6 cos (x) = 2

Let:

cos (x) = u

- 4 u^{2} + 6 u = 2

- 4 u^{2} + 6 u = 2 : u = \frac{1}{2}, u = 1

- 4 u^{2} + 6 u = 2

Move

2

to the left side

- 4 u^{2} + 6 u = 2

Subtract

2

from both sides

- 4 u^{2} + 6 u - 2 = 2 - 2

Simplify

- 4 u^{2} + 6 u - 2 = 0

- 4 u^{2} + 6 u - 2 = 0

Solve with the quadratic formula

- 4 u^{2} + 6 u - 2 = 0

Quadratic Equation Formula:

For

a = - 4, b = 6, c = - 2

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 4 ) ( - 2 )}{2 ( - 4 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 4 ) ( - 2 )}{2 ( - 4 )}

6^{2} - 4 (- 4) (- 2) = 2

6^{2} - 4 (- 4) (- 2)

Apply rule

- (- a) = a

= 6^{2} - 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

u_{1, 2} = \frac{- 6 \pm 2}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 6 + 2}{2 ( - 4 )}, u_{2} = \frac{- 6 - 2}{2 ( - 4 )}

u = \frac{- 6 + 2}{2 ( - 4 )} : \frac{1}{2}

\frac{- 6 + 2}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 2}{- 2 \cdot 4}

Add/Subtract the numbers:

- 6 + 2 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{- 6 - 2}{2 ( - 4 )} : 1

\frac{- 6 - 2}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 2}{- 2 \cdot 4}

Subtract the numbers:

- 6 - 2 = - 8

= \frac{- 8}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{- 8}

Apply the fraction rule:

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

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -4cos^2(x)+6cos(x)=2 ?
The general solution for -4cos^2(x)+6cos(x)=2 is x= pi/3+2pin,x=(5pi)/3+2pin,x=2pin