What is the general solution for cos^2(x)+5cos(x)=cos(x)-3 ?

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

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cos^2(x)+5cos(x)=cos(x)-3

Solution

cos^{2} (x) + 5 cos (x) = cos (x) - 3

Solution

x = π + 2 πn

Degrees

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

Solution steps

cos^{2} (x) + 5 cos (x) = cos (x) - 3

Solve by substitution

cos^{2} (x) + 5 cos (x) = cos (x) - 3

Let:

cos (x) = u

u^{2} + 5 u = u - 3

u^{2} + 5 u = u - 3 : u = - 1, u = - 3

u^{2} + 5 u = u - 3

Move

3

to the left side

u^{2} + 5 u = u - 3

Add

3

to both sides

u^{2} + 5 u + 3 = u - 3 + 3

Simplify

u^{2} + 5 u + 3 = u

u^{2} + 5 u + 3 = u

Move

u

to the left side

u^{2} + 5 u + 3 = u

Subtract

u

from both sides

u^{2} + 5 u + 3 - u = u - u

Simplify

u^{2} + 4 u + 3 = 0

u^{2} + 4 u + 3 = 0

Solve with the quadratic formula

u^{2} + 4 u + 3 = 0

Quadratic Equation Formula:

For

a = 1, b = 4, c = 3

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

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

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

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

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

u_{1} = \frac{- 4 + 2}{2 \cdot 1}, u_{2} = \frac{- 4 - 2}{2 \cdot 1}

u = \frac{- 4 + 2}{2 \cdot 1} : - 1

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

Add/Subtract the numbers:

- 4 + 2 = - 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

u = \frac{- 4 - 2}{2 \cdot 1} : - 3

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

Subtract the numbers:

- 4 - 2 = - 6

= \frac{- 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{2}

Apply the fraction rule:

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

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

The solutions to the quadratic equation are:

u = - 1, u = - 3

Substitute back

u = cos (x)

cos (x) = - 1, cos (x) = - 3

cos (x) = - 1, cos (x) = - 3

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) = - 3 :

No Solution

cos (x) = - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = π + 2 πn

Popular Examples

-9sin(x)-9cos(x)=0

- 9 sin (x) - 9 cos (x) = 0

cot(x)= 1/3

cot (x) = \frac{1}{3}

sin(x)= 5/2

sin (x) = \frac{5}{2}

2sin^2(θ)+sqrt(2)sin(θ)=0

2 sin^{2} (θ) + 2 sin (θ) = 0

cos(3.7x+5.2)=0

cos (3.7 x + 5.2) = 0

Frequently Asked Questions (FAQ)

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