What is the general solution for cos^2(x)-4cos(x)=0 ?

The general solution for cos^2(x)-4cos(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(x)-4cos(x)=0

Solution

cos^{2} (x) - 4 cos (x) = 0

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^{2} (x) - 4 cos (x) = 0

Solve by substitution

cos^{2} (x) - 4 cos (x) = 0

Let:

cos (x) = u

u^{2} - 4 u = 0

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

u^{2} - 4 u = 0

Solve with the quadratic formula

u^{2} - 4 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 0

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 0 = 4

(- 4)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

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

n

is even

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

= 4^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} - 0

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

= 4^{2}

Apply radical rule:

assuming

a \geq 0

= 4

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

4 + 4 = 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{- ( - 4 ) - 4}{2 \cdot 1} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

4 - 4 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 4, u = 0

Substitute back

u = cos (x)

cos (x) = 4, cos (x) = 0

cos (x) = 4, cos (x) = 0

cos (x) = 4 :

No Solution

cos (x) = 4

- 1 \leq cos (x) \leq 1

No Solution

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

Combine all the solutions

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

Popular Examples

solvefor x,sin(x)+cot(x)cos(x)= 1/f solve for

x, sin (x) + cot (x) cos (x) = \frac{1}{f}

1/(sin(x))=2

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

tan(x)csc(x)-2tan(x)=0

tan (x) csc (x) - 2 tan (x) = 0

1-2sin(θ)=0

1 - 2 sin (θ) = 0

cos(2θ)+3cos(θ)=4

cos (2 θ) + 3 cos (θ) = 4

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-4cos(x)=0 ?
The general solution for cos^2(x)-4cos(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin