What is the general solution for cos^2(x)+6cos(x)=-3,y=3 ?

The general solution for cos^2(x)+6cos(x)=-3,y=3 is No Solution for x\in\mathbb{R}

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cos^2(x)+6cos(x)=-3,y=3

Solution

cos^{2} (x) + 6 cos (x) = - 3, y = 3

Solution

No Solution for x \in R

Solution steps

cos^{2} (x) + 6 cos (x) = - 3, y = 3

Solve by substitution

cos^{2} (x) + 6 cos (x) = - 3

Let:

cos (x) = u

u^{2} + 6 u = - 3

u^{2} + 6 u = - 3 : u = - 3 + 6, u = - 3 - 6

u^{2} + 6 u = - 3

Move

3

to the left side

u^{2} + 6 u = - 3

Add

3

to both sides

u^{2} + 6 u + 3 = - 3 + 3

Simplify

u^{2} + 6 u + 3 = 0

u^{2} + 6 u + 3 = 0

Solve with the quadratic formula

u^{2} + 6 u + 3 = 0

Quadratic Equation Formula:

For

a = 1, b = 6, c = 3

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 3 = 26

6^{2} - 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 6^{2} - 12

6^{2} = 36

= 36 - 12

Subtract the numbers:

36 - 12 = 24

= 24

Prime factorization of

24 : 2^{3} \cdot 3

24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3

= 2^{3} \cdot 3

= 2^{3} \cdot 3

Apply exponent rule:

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

= 2^{2} \cdot 2 \cdot 3

Apply radical rule:

= 2^{2} 2 \cdot 3

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 3

Refine

= 26

u_{1, 2} = \frac{- 6 \pm 2 6}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 6 + 2 6}{2 \cdot 1}, u_{2} = \frac{- 6 - 2 6}{2 \cdot 1}

u = \frac{- 6 + 2 6}{2 \cdot 1} : - 3 + 6

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6 + 2 6}{2}

Factor

- 6 + 26 : 2 (- 3 + 6)

- 6 + 26

Rewrite as

= - 2 \cdot 3 + 26

Factor out common term

2

= 2 (- 3 + 6)

= \frac{2 ( - 3 + 6 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3 + 6

u = \frac{- 6 - 2 6}{2 \cdot 1} : - 3 - 6

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 6 - 26 : - 2 (3 + 6)

- 6 - 26

Rewrite as

= - 2 \cdot 3 - 26

Factor out common term

2

= - 2 (3 + 6)

= - \frac{2 ( 3 + 6 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (3 + 6)

Negate

- (3 + 6) = - 3 - 6

= - 3 - 6

The solutions to the quadratic equation are:

u = - 3 + 6, u = - 3 - 6

Substitute back

u = cos (x)

cos (x) = - 3 + 6, cos (x) = - 3 - 6

cos (x) = - 3 + 6, cos (x) = - 3 - 6

cos (x) = - 3 + 6, y = 3 :

No Solution

cos (x) = - 3 + 6, y = 3

Apply trig inverse properties

cos (x) = - 3 + 6

General solutions for

cos (x) = - 3 + 6

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- 3 + 6) + 2 πn, x = - arccos (- 3 + 6) + 2 πn

x = arccos (- 3 + 6) + 2 πn, x = - arccos (- 3 + 6) + 2 πn

Solutions for the range

y = 3

No Solution

cos (x) = - 3 - 6, y = 3 :

No Solution

cos (x) = - 3 - 6, y = 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)+6cos(x)=-3,y=3 ?
The general solution for cos^2(x)+6cos(x)=-3,y=3 is No Solution for x\in\mathbb{R}