What is the general solution for 1-2cos^2(x)-cos(x)=0 ?

The general solution for 1-2cos^2(x)-cos(x)=0 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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1-2cos^2(x)-cos(x)=0

Solution

1 - 2 cos^{2} (x) - cos (x) = 0

Solution

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

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

1 - 2 cos^{2} (x) - cos (x) = 0

Solve by substitution

1 - 2 cos^{2} (x) - cos (x) = 0

Let:

cos (x) = u

1 - 2 u^{2} - u = 0

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

1 - 2 u^{2} - u = 0

Write in the standard form

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

- 2 u^{2} - u + 1 = 0

Solve with the quadratic formula

- 2 u^{2} - u + 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 2) \cdot 1 = 3

(- 1)^{2} - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + 3}{- 2 \cdot 2}

Add the numbers:

1 + 3 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

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

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - 3}{- 2 \cdot 2}

Subtract the numbers:

1 - 3 = - 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 = - 1, u = \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

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

Combine all the solutions

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

Popular Examples

1/3 =cos((pix)/(12))

\frac{1}{3} = cos (\frac{π x}{12})

(cos(30))/(cos(60))=tan(x)

\frac{cos ( 3 0 ^{\circ} )}{cos ( 6 0 ^{\circ} )} = tan (x)

4cos^2(x)-18cos(x)-10=0

4 cos^{2} (x) - 18 cos (x) - 10 = 0

25sin(θ)-1.5cos(θ)=20

25 sin (θ) - 1.5 cos (θ) = 20

sin((2θ)/3)= 1/2 ,0<= θ<= 2pi

sin (\frac{2 θ}{3}) = \frac{1}{2}, 0 \leq θ \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 1-2cos^2(x)-cos(x)=0 ?
The general solution for 1-2cos^2(x)-cos(x)=0 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin