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

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

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

Solution

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- 2 u^{2} = u - 1

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

- 2 u^{2} = u - 1

Move

1

to the left side

- 2 u^{2} = u - 1

Add

1

to both sides

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

Simplify

- 2 u^{2} + 1 = u

- 2 u^{2} + 1 = u

Move

u

to the left side

- 2 u^{2} + 1 = u

Subtract

u

from both sides

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

Simplify

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

- 2 u^{2} + 1 - 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

cos(2x)-sin(2x)=0

cos (2 x) - sin (2 x) = 0

cos(5θ)=0

cos (5 θ) = 0

sin(θ)=csc(θ)

sin (θ) = csc (θ)

-5sin(x)=0,0<= x<= 2pi

- 5 sin (x) = 0, 0 \leq x \leq 2 π

3sec^2(θ)-4=0

3 sec^{2} (θ) - 4 = 0

Frequently Asked Questions (FAQ)

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