What is the general solution for cos^3(x)-3cos(x)=3cos(x)sin(x) ?

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

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cos^3(x)-3cos(x)=3cos(x)sin(x)

Solution

cos^{3} (x) - 3 cos (x) = 3 cos (x) sin (x)

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{3} (x) - 3 cos (x) = 3 cos (x) sin (x)

Subtract

3 cos (x) sin (x)

from both sides

cos^{3} (x) - 3 cos (x) - 3 cos (x) sin (x) = 0

Factor

cos^{3} (x) - 3 cos (x) - 3 cos (x) sin (x) : cos (x) (cos^{2} (x) - 3 - 3 sin (x))

cos^{3} (x) - 3 cos (x) - 3 cos (x) sin (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

cos^{3} (x) = cos (x) cos^{2} (x)

= cos (x) cos^{2} (x) - 3 cos (x) - 3 sin (x) cos (x)

Factor out common term

cos (x)

= cos (x) (cos^{2} (x) - 3 - 3 sin (x))

cos (x) (cos^{2} (x) - 3 - 3 sin (x)) = 0

Solving each part separately

cos (x) = 0 or cos^{2} (x) - 3 - 3 sin (x) = 0

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

cos (x) = 0

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

cos^{2} (x) - 3 - 3 sin (x) = 0 : x = \frac{3 π}{2} + 2 πn

cos^{2} (x) - 3 - 3 sin (x) = 0

Rewrite using trig identities

- 3 + cos^{2} (x) - 3 sin (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= - 3 + 1 - sin^{2} (x) - 3 sin (x)

Simplify

= - sin^{2} (x) - 3 sin (x) - 2

- 2 - sin^{2} (x) - 3 sin (x) = 0

Solve by substitution

- 2 - sin^{2} (x) - 3 sin (x) = 0

Let:

sin (x) = u

- 2 - u^{2} - 3 u = 0

- 2 - u^{2} - 3 u = 0 : u = - 2, u = - 1

- 2 - u^{2} - 3 u = 0

Write in the standard form

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

- u^{2} - 3 u - 2 = 0

Solve with the quadratic formula

- u^{2} - 3 u - 2 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 3, c = - 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

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

(- 3)^{2} - 4 (- 1) (- 2)

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

3 + 1 = 4

= \frac{4}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{- 2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

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

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

Remove parentheses:

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

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

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - 2, u = - 1

Substitute back

u = sin (x)

sin (x) = - 2, sin (x) = - 1

sin (x) = - 2, sin (x) = - 1

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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