What is the general solution for 5cos^2(x)-12sin^2(x)=13 ?

The general solution for 5cos^2(x)-12sin^2(x)=13 is No Solution for x\in\mathbb{R}

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Popular >

5cos^2(x)-12sin^2(x)=13

Solution

5 cos^{2} (x) - 12 sin^{2} (x) = 13

Solution

No Solution for x \in R

Solution steps

5 cos^{2} (x) - 12 sin^{2} (x) = 13

Subtract

13

from both sides

5 cos^{2} (x) - 12 sin^{2} (x) - 13 = 0

Rewrite using trig identities

- 13 - 12 sin^{2} (x) + 5 cos^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 13 - 12 sin^{2} (x) + 5 (1 - sin^{2} (x))

Simplify

- 13 - 12 sin^{2} (x) + 5 (1 - sin^{2} (x)) : - 17 sin^{2} (x) - 8

- 13 - 12 sin^{2} (x) + 5 (1 - sin^{2} (x))

Expand

5 (1 - sin^{2} (x)) : 5 - 5 sin^{2} (x)

5 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 1, c = sin^{2} (x)

= 5 \cdot 1 - 5 sin^{2} (x)

Multiply the numbers:

5 \cdot 1 = 5

= 5 - 5 sin^{2} (x)

= - 13 - 12 sin^{2} (x) + 5 - 5 sin^{2} (x)

Simplify

- 13 - 12 sin^{2} (x) + 5 - 5 sin^{2} (x) : - 17 sin^{2} (x) - 8

- 13 - 12 sin^{2} (x) + 5 - 5 sin^{2} (x)

Group like terms

= - 12 sin^{2} (x) - 5 sin^{2} (x) - 13 + 5

Add similar elements:

- 12 sin^{2} (x) - 5 sin^{2} (x) = - 17 sin^{2} (x)

= - 17 sin^{2} (x) - 13 + 5

Add/Subtract the numbers:

- 13 + 5 = - 8

= - 17 sin^{2} (x) - 8

= - 17 sin^{2} (x) - 8

= - 17 sin^{2} (x) - 8

- 8 - 17 sin^{2} (x) = 0

Solve by substitution

- 8 - 17 sin^{2} (x) = 0

Let:

sin (x) = u

- 8 - 17 u^{2} = 0

- 8 - 17 u^{2} = 0 : u = i \frac{2 34}{17}, u = - i \frac{2 34}{17}

- 8 - 17 u^{2} = 0

Move

8

to the right side

- 8 - 17 u^{2} = 0

Add

8

to both sides

- 8 - 17 u^{2} + 8 = 0 + 8

Simplify

- 17 u^{2} = 8

- 17 u^{2} = 8

Divide both sides by

- 17

- 17 u^{2} = 8

Divide both sides by

- 17

\frac{- 17 u ^{2}}{- 17} = \frac{8}{- 17}

Simplify

u^{2} = - \frac{8}{17}

u^{2} = - \frac{8}{17}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - \frac{8}{17}, u = - - \frac{8}{17}

Simplify

- \frac{8}{17} : i \frac{2 34}{17}

- \frac{8}{17}

Apply radical rule:

- a = - 1 a

- \frac{8}{17} = - 1 \frac{8}{17}

= - 1 \frac{8}{17}

Apply imaginary number rule:

- 1 = i

= i \frac{8}{17}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{8}{17} = \frac{8}{17}

= i \frac{8}{17}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= i \frac{2 2}{17}

\frac{2 2}{17} = \frac{2 34}{17}

\frac{2 2}{17}

Multiply by the conjugate

\frac{17}{17}

= \frac{2 2 17}{17 17}

2217 = 234

2217

Apply radical rule:

a b = a \cdot b

217 = 2 \cdot 17

= 2 2 \cdot 17

Multiply the numbers:

2 \cdot 17 = 34

= 234

1717 = 17

1717

Apply radical rule:

a a = a

1717 = 17

= 17

= \frac{2 34}{17}

= i \frac{2 34}{17}

Rewrite

i \frac{2 34}{17}

in standard complex form:

\frac{2 34}{17} i

i \frac{2 34}{17}

\frac{2 34}{17} = \frac{2 2}{17}

\frac{2 34}{17}

Factor

34 : 217

Factor

34 = 2 \cdot 17

= 2 \cdot 17

Apply radical rule:

= 217

= \frac{2 2 17}{17}

Cancel

\frac{2 2 17}{17} : \frac{2 2}{17}

\frac{2 2 17}{17}

Apply radical rule:

17 = 1 7^{\frac{1}{2}}

= \frac{2 2 \cdot 1 7 ^{\frac{1}{2}}}{17}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

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

Subtract the numbers:

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

= \frac{2 2}{1 7 ^{\frac{1}{2}}}

Apply radical rule:

1 7^{\frac{1}{2}} = 17

= \frac{2 2}{17}

= \frac{2 2}{17}

= i \frac{2 2}{17}

Multiply fractions:

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

= \frac{2 2 i}{17}

\frac{2 2}{17} = \frac{2 34}{17}

\frac{2 2}{17}

Multiply by the conjugate

\frac{17}{17}

= \frac{2 2 17}{17 17}

2217 = 234

2217

Apply radical rule:

a b = a \cdot b

217 = 2 \cdot 17

= 2 2 \cdot 17

Multiply the numbers:

2 \cdot 17 = 34

= 234

1717 = 17

1717

Apply radical rule:

a a = a

1717 = 17

= 17

= \frac{2 34}{17}

= \frac{2 34}{17} i

= \frac{2 34}{17} i

Simplify

- - \frac{8}{17} : - i \frac{2 34}{17}

- - \frac{8}{17}

Simplify

- \frac{8}{17} : i \frac{2 2}{17}

- \frac{8}{17}

Apply radical rule:

- a = - 1 a

- \frac{8}{17} = - 1 \frac{8}{17}

= - 1 \frac{8}{17}

Apply imaginary number rule:

- 1 = i

= i \frac{8}{17}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{8}{17} = \frac{8}{17}

= i \frac{8}{17}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= i \frac{2 2}{17}

= - i \frac{2 2}{17}

\frac{2 2}{17} = \frac{2 34}{17}

\frac{2 2}{17}

Multiply by the conjugate

\frac{17}{17}

= \frac{2 2 17}{17 17}

2217 = 234

2217

Apply radical rule:

a b = a \cdot b

217 = 2 \cdot 17

= 2 2 \cdot 17

Multiply the numbers:

2 \cdot 17 = 34

= 234

1717 = 17

1717

Apply radical rule:

a a = a

1717 = 17

= 17

= \frac{2 34}{17}

= - \frac{2 34}{17} i

u = i \frac{2 34}{17}, u = - i \frac{2 34}{17}

Substitute back

u = sin (x)

sin (x) = i \frac{2 34}{17}, sin (x) = - i \frac{2 34}{17}

sin (x) = i \frac{2 34}{17}, sin (x) = - i \frac{2 34}{17}

sin (x) = i \frac{2 34}{17} :

No Solution

sin (x) = i \frac{2 34}{17}

No Solution

sin (x) = - i \frac{2 34}{17} :

No Solution

sin (x) = - i \frac{2 34}{17}

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 5cos^2(x)-12sin^2(x)=13 ?
The general solution for 5cos^2(x)-12sin^2(x)=13 is No Solution for x\in\mathbb{R}