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

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

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

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

2 u^{2} + u = 0

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

2 u^{2} + u = 0

Solve with the quadratic formula

2 u^{2} + u = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1^{2} - 4 \cdot 2 \cdot 0 = 1

1^{2} - 4 \cdot 2 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- 1 + 1}{2 \cdot 2} : 0

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

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

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

Subtract the numbers:

- 1 - 1 = - 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}

The solutions to the quadratic equation are:

u = 0, u = - \frac{1}{2}

Substitute back

u = cos (x)

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

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

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

cos (x) = 0, 0 \leq x \leq 2 π

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 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{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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

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

Popular Examples

sin^2(x)-2cos(x)=2

sin^{2} (x) - 2 cos (x) = 2

2cos(θ)-sin(θ)=0

2 cos (θ) - sin (θ) = 0

cos(x+pi/6)=sin(x)

cos (x + \frac{π}{6}) = sin (x)

sin(θ)=-1/5

sin (θ) = - \frac{1}{5}

5cot(x)-5=0

5 cot (x) - 5 = 0

Frequently Asked Questions (FAQ)

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