What is the general solution for cos^2(x)-7cos(x)+12=0 ?

The general solution for cos^2(x)-7cos(x)+12=0 is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(x)-7cos(x)+12=0

Solution

cos^{2} (x) - 7 cos (x) + 12 = 0

Solution

No Solution for x \in R

Solution steps

cos^{2} (x) - 7 cos (x) + 12 = 0

Solve by substitution

cos^{2} (x) - 7 cos (x) + 12 = 0

Let:

cos (x) = u

u^{2} - 7 u + 12 = 0

u^{2} - 7 u + 12 = 0 : u = 4, u = 3

u^{2} - 7 u + 12 = 0

Solve with the quadratic formula

u^{2} - 7 u + 12 = 0

Quadratic Equation Formula:

For

a = 1, b = - 7, c = 12

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 12 = 1

(- 7)^{2} - 4 \cdot 1 \cdot 12

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 1 \cdot 12

Multiply the numbers:

4 \cdot 1 \cdot 12 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

Subtract the numbers:

49 - 48 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 1}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 1}

u = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

\frac{- ( - 7 ) + 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 + 1}{2 \cdot 1}

Add the numbers:

7 + 1 = 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{- ( - 7 ) - 1}{2 \cdot 1} : 3

\frac{- ( - 7 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 - 1}{2 \cdot 1}

Subtract the numbers:

7 - 1 = 6

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

The solutions to the quadratic equation are:

u = 4, u = 3

Substitute back

u = cos (x)

cos (x) = 4, cos (x) = 3

cos (x) = 4, cos (x) = 3

cos (x) = 4 :

No Solution

cos (x) = 4

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 3 :

No Solution

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

Graph

Popular Examples

2sin^2(x)+sin(x)-1=0,0<= x<= 2pi

2 sin^{2} (x) + sin (x) - 1 = 0, 0 \leq x \leq 2 π

2sin(e^{t/4})+1=0

2 sin (e^{\frac{t}{4}}) + 1 = 0

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

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

sin(u)=-2/5

sin (u) = - \frac{2}{5}

2sin(3x)cos(2x)-cos(2x)=0

2 sin (3 x) cos (2 x) - cos (2 x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-7cos(x)+12=0 ?
The general solution for cos^2(x)-7cos(x)+12=0 is No Solution for x\in\mathbb{R}