What is the general solution for cos^2(x)-3cos(x)=4 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

cos^2(x)-3cos(x)=4

Solution

cos^{2} (x) - 3 cos (x) = 4

Solution

x = π + 2 πn

Degrees

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

Solution steps

cos^{2} (x) - 3 cos (x) = 4

Solve by substitution

cos^{2} (x) - 3 cos (x) = 4

Let:

cos (x) = u

u^{2} - 3 u = 4

u^{2} - 3 u = 4 : u = 4, u = - 1

u^{2} - 3 u = 4

Move

4

to the left side

u^{2} - 3 u = 4

Subtract

4

from both sides

u^{2} - 3 u - 4 = 4 - 4

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 3, c = - 4

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

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

u = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

\frac{- ( - 3 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 1}

Add the numbers:

3 + 5 = 8

= \frac{8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

u = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

\frac{- ( - 3 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 1}

Subtract the numbers:

3 - 5 = - 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 = 4, u = - 1

Substitute back

u = cos (x)

cos (x) = 4, cos (x) = - 1

cos (x) = 4, cos (x) = - 1

cos (x) = 4 :

No Solution

cos (x) = 4

- 1 \leq cos (x) \leq 1

No Solution

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

Combine all the solutions

x = π + 2 πn

Graph

Popular Examples

csc(θ)= 5/2

solvefor x,sin^2(x)= 1/2

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

3sqrt(2)cos(θ)+2=5

cos^2(x)+3sin(x)=3

Frequently Asked Questions (FAQ)

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