What is the general solution for 4cos(x)=cos^3(x) ?

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

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4cos(x)=cos^3(x)

Solution

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

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

Solve by substitution

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

Let:

cos (x) = u

4 u = u^{3}

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

4 u = u^{3}

Switch sides

u^{3} = 4 u

Move

4 u

to the left side

u^{3} = 4 u

Subtract

4 u

from both sides

u^{3} - 4 u = 4 u - 4 u

Simplify

u^{3} - 4 u = 0

u^{3} - 4 u = 0

Factor

u^{3} - 4 u : u (u + 2) (u - 2)

u^{3} - 4 u

Factor out common term

u : u (u^{2} - 4)

u^{3} - 4 u

Apply exponent rule:

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

u^{3} = u^{2} u

= u^{2} u - 4 u

Factor out common term

u

= u (u^{2} - 4)

= u (u^{2} - 4)

Factor

u^{2} - 4 : (u + 2) (u - 2)

u^{2} - 4

Rewrite

4

2^{2}

= u^{2} - 2^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 2^{2} = (u + 2) (u - 2)

= (u + 2) (u - 2)

= u (u + 2) (u - 2)

u (u + 2) (u - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u + 2 = 0 or u - 2 = 0

Solve

u + 2 = 0 : u = - 2

u + 2 = 0

Move

2

to the right side

u + 2 = 0

Subtract

2

from both sides

u + 2 - 2 = 0 - 2

Simplify

u = - 2

u = - 2

Solve

u - 2 = 0 : u = 2

u - 2 = 0

Move

2

to the right side

u - 2 = 0

Add

2

to both sides

u - 2 + 2 = 0 + 2

Simplify

u = 2

u = 2

The solutions are

u = 0, u = - 2, u = 2

Substitute back

u = cos (x)

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

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

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 (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Popular Examples

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cos^{2} (a) = \frac{3}{5 sin ( a )}

3cos(x)=2sec(x)-5

3 cos (x) = 2 sec (x) - 5

sin(a/2)=(1/2)sin(a)

sin (\frac{a}{2}) = (\frac{1}{2}) sin (a)

cos(x/2)=cos(x)+1

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

d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x))

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 \cdot cos ^{2} ( x )}

Frequently Asked Questions (FAQ)

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