What is the general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi ?

Q: What is the general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi ?

The general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=pi

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4cos^2(x)+2cos(x)=2,0<= x<= 2pi

Solution

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

Solution

x = \frac{π}{3}, x = \frac{5 π}{3}, x = π

Degrees

x = 6 0^{\circ}, x = 30 0^{\circ}, x = 18 0^{\circ}

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

4 u^{2} + 2 u = 2

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

4 u^{2} + 2 u = 2

Move

2

to the left side

4 u^{2} + 2 u = 2

Subtract

2

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 6 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

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

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

Subtract the numbers:

- 2 - 6 = - 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}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 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

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{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 = \frac{π}{3}, x = \frac{5 π}{3}, x = π

Popular Examples

arctan(x)+arctan(2x)= pi/4

arctan (x) + arctan (2 x) = \frac{π}{4}

1+cos(θ)=(2+sqrt(3))/2

1 + cos (θ) = \frac{2 + 3}{2}

1+3sin(x)=2

1 + 3 sin (x) = 2

cos(x/2+pi/3)= 1/(sqrt(2))

cos (\frac{x}{2} + \frac{π}{3}) = \frac{1}{2}

sinh(x)=(sqrt(2))/2

sinh (x) = \frac{2}{2}

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi ?
The general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=pi