What is the general solution for cos(x)(18-12sqrt(2))=12+6sqrt(3) ?

The general solution for cos(x)(18-12sqrt(2))=12+6sqrt(3) is No Solution for x\in\mathbb{R}

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cos(x)(18-12sqrt(2))=12+6sqrt(3)

Solution

cos (x) (18 - 122) = 12 + 63

Solution

No Solution for x \in R

Solution steps

cos (x) (18 - 122) = 12 + 63

Divide both sides by

18 - 122

cos (x) (18 - 122) = 12 + 63

Divide both sides by

18 - 122

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2} = \frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2}

Simplify

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2} = \frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2}

Simplify

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2} : cos (x)

\frac{cos ( x ) ( 18 - 12 2 )}{18 - 12 2}

Cancel the common factor:

18 - 122

= cos (x)

Simplify

\frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2} : (2 + 3) (3 + 22)

\frac{12}{18 - 12 2} + \frac{6 3}{18 - 12 2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{12 + 6 3}{18 - 12 2}

Factor

12 + 63 : 6 (2 + 3)

12 + 63

Rewrite as

= 6 \cdot 2 + 63

Factor out common term

6

= 6 (2 + 3)

= \frac{6 ( 2 + 3 )}{18 - 12 2}

Factor

18 - 122 : 6 (3 - 22)

18 - 122

Rewrite as

= 6 \cdot 3 - 6 \cdot 22

Factor out common term

6

= 6 (3 - 22)

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

Divide the numbers:

\frac{6}{6} = 1

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

Remove parentheses:

(a) = a

= \frac{2 + 3}{3 - 2 2}

Rationalize

\frac{2 + 3}{3 - 2 2} : (2 + 3) (3 + 22)

\frac{2 + 3}{3 - 2 2}

Multiply by the conjugate

\frac{3 + 2 2}{3 + 2 2}

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

(3 - 22) (3 + 22) = 1

(3 - 22) (3 + 22)

22 = 2^{\frac{3}{2}}

22

Apply exponent rule:

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

22 = 2 \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2}}

= 2^{1 + \frac{1}{2}}

Join

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= 2^{\frac{3}{2}}

= (3 - 2^{\frac{3}{2}}) (3 + 22)

22 = 2^{\frac{3}{2}}

22

Apply exponent rule:

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

22 = 2 \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2}}

= 2^{1 + \frac{1}{2}}

Join

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= 2^{\frac{3}{2}}

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

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = 3, b = 2^{\frac{3}{2}}

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

Simplify

3^{2} - (2^{\frac{3}{2}})^{2} : 1

3^{2} - (2^{\frac{3}{2}})^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

(2^{\frac{3}{2}})^{2} = 8

(2^{\frac{3}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

\frac{3}{2} \cdot 2 = 3

\frac{3}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 \cdot 2}{2}

Cancel the common factor:

2

= 3

= 2^{3}

2^{3} = 8

= 8

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

= 1

= \frac{( 2 + 3 ) ( 3 + 2 2 )}{1}

Apply rule

\frac{a}{1} = a

= (2 + 3) (3 + 22)

= (2 + 3) (3 + 22)

cos (x) = (2 + 3) (3 + 22)

cos (x) = (2 + 3) (3 + 22)

cos (x) = (2 + 3) (3 + 22)

- 1 \leq cos (x) \leq 1

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)(18-12sqrt(2))=12+6sqrt(3) ?
The general solution for cos(x)(18-12sqrt(2))=12+6sqrt(3) is No Solution for x\in\mathbb{R}