What is the general solution for -sin(x)-2cot(x)+5csc(x)=0 ?

The general solution for -sin(x)-2cot(x)+5csc(x)=0 is No Solution for x\in\mathbb{R}

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Popular >

-sin(x)-2cot(x)+5csc(x)=0

Solution

- sin (x) - 2 cot (x) + 5 csc (x) = 0

Solution

No Solution for x \in R

Solution steps

- sin (x) - 2 cot (x) + 5 csc (x) = 0

Express with sin, cos

- sin (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} + 5 \cdot \frac{1}{sin ( x )} = 0

Simplify

- sin (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} + 5 \cdot \frac{1}{sin ( x )} : \frac{- sin ^{2} ( x ) - 2 cos ( x ) + 5}{sin ( x )}

- sin (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} + 5 \cdot \frac{1}{sin ( x )}

2 \cdot \frac{cos ( x )}{sin ( x )} = \frac{2 cos ( x )}{sin ( x )}

2 \cdot \frac{cos ( x )}{sin ( x )}

Multiply fractions:

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

= \frac{cos ( x ) \cdot 2}{sin ( x )}

5 \cdot \frac{1}{sin ( x )} = \frac{5}{sin ( x )}

5 \cdot \frac{1}{sin ( x )}

Multiply fractions:

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

= \frac{1 \cdot 5}{sin ( x )}

Multiply the numbers:

1 \cdot 5 = 5

= \frac{5}{sin ( x )}

= - sin (x) - \frac{2 cos ( x )}{sin ( x )} + \frac{5}{sin ( x )}

Combine the fractions

- \frac{2 cos ( x )}{sin ( x )} + \frac{5}{sin ( x )} : \frac{- 2 cos ( x ) + 5}{sin ( x )}

Apply rule

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

= \frac{- 2 cos ( x ) + 5}{sin ( x )}

= - sin (x) + \frac{- 2 cos ( x ) + 5}{sin ( x )}

Convert element to fraction:

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

= - \frac{sin ( x ) sin ( x )}{sin ( x )} + \frac{- cos ( x ) \cdot 2 + 5}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( x ) sin ( x ) - cos ( x ) \cdot 2 + 5}{sin ( x )}

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

- sin (x) sin (x) - cos (x) \cdot 2 + 5

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

= \frac{- sin ^{2} ( x ) - 2 cos ( x ) + 5}{sin ( x )}

\frac{- sin ^{2} ( x ) - 2 cos ( x ) + 5}{sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Add

2 cos (x)

to both sides

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

Square both sides

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

Subtract

(2 cos (x))^{2}

from both sides

(- sin^{2} (x) + 5)^{2} - 4 cos^{2} (x) = 0

Factor

(- sin^{2} (x) + 5)^{2} - 4 cos^{2} (x) : (- sin^{2} (x) + 5 + 2 cos (x)) (- sin^{2} (x) + 5 - 2 cos (x))

(- sin^{2} (x) + 5)^{2} - 4 cos^{2} (x)

Rewrite

(- sin^{2} (x) + 5)^{2} - 4 cos^{2} (x)

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

(- sin^{2} (x) + 5)^{2} - 4 cos^{2} (x)

Rewrite

4

2^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

2^{2} cos^{2} (x) = (2 cos (x))^{2}

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Refine

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

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

Solving each part separately

- sin^{2} (x) + 5 + 2 cos (x) = 0 or - sin^{2} (x) + 5 - 2 cos (x) = 0

- sin^{2} (x) + 5 + 2 cos (x) = 0 :

No Solution

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 5 - (1 - cos^{2} (x)) + 2 cos (x)

Simplify

5 - (1 - cos^{2} (x)) + 2 cos (x) : cos^{2} (x) + 2 cos (x) + 4

5 - (1 - cos^{2} (x)) + 2 cos (x)

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

= 5 - 1 + cos^{2} (x) + 2 cos (x)

Subtract the numbers:

5 - 1 = 4

= cos^{2} (x) + 2 cos (x) + 4

= cos^{2} (x) + 2 cos (x) + 4

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

Solve by substitution

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

Let:

cos (x) = u

4 + u^{2} + 2 u = 0

4 + u^{2} + 2 u = 0 : u = - 1 + 3 i, u = - 1 - 3 i

4 + u^{2} + 2 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + 2 u + 4 = 0

Solve with the quadratic formula

u^{2} + 2 u + 4 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 4

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

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

Simplify

2^{2} - 4 \cdot 1 \cdot 4 : 23 i

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 2^{2} - 16

Apply imaginary number rule:

- a = i a

= i 16 - 2^{2}

- 2^{2} + 16 = 23

- 2^{2} + 16

2^{2} = 4

= - 4 + 16

Add/Subtract the numbers:

- 4 + 16 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

= 23 i

u_{1, 2} = \frac{- 2 \pm 2 3 i}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 2 + 2 3 i}{2 \cdot 1}, u_{2} = \frac{- 2 - 2 3 i}{2 \cdot 1}

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

\frac{- 2 + 2 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 + 23 i : 2 (- 1 + 3 i)

- 2 + 23 i

Rewrite as

= - 2 \cdot 1 + 23 i

Factor out common term

2

= 2 (- 1 + 3 i)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 3 i

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

\frac{- 2 - 2 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 - 23 i : - 2 (1 + 3 i)

- 2 - 23 i

Rewrite as

= - 2 \cdot 1 - 23 i

Factor out common term

2

= - 2 (1 + 3 i)

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

Divide the numbers:

\frac{2}{2} = 1

= - (1 + 3 i)

Negate

- (1 + 3 i) = - 1 - 3 i

= - 1 - 3 i

The solutions to the quadratic equation are:

u = - 1 + 3 i, u = - 1 - 3 i

Substitute back

u = cos (x)

cos (x) = - 1 + 3 i, cos (x) = - 1 - 3 i

cos (x) = - 1 + 3 i, cos (x) = - 1 - 3 i

cos (x) = - 1 + 3 i :

No Solution

cos (x) = - 1 + 3 i

No Solution

cos (x) = - 1 - 3 i :

No Solution

cos (x) = - 1 - 3 i

No Solution

Combine all the solutions

No Solution

- sin^{2} (x) + 5 - 2 cos (x) = 0 :

No Solution

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 5 - (1 - cos^{2} (x)) - 2 cos (x)

Simplify

5 - (1 - cos^{2} (x)) - 2 cos (x) : cos^{2} (x) - 2 cos (x) + 4

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

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

= 5 - 1 + cos^{2} (x) - 2 cos (x)

Subtract the numbers:

5 - 1 = 4

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

4 + u^{2} - 2 u = 0 : u = 1 + 3 i, u = 1 - 3 i

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 4

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

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 4 : 23 i

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 2^{2} - 16

Apply imaginary number rule:

- a = i a

= i 16 - 2^{2}

- 2^{2} + 16 = 23

- 2^{2} + 16

2^{2} = 4

= - 4 + 16

Add/Subtract the numbers:

- 4 + 16 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

= 23 i

u_{1, 2} = \frac{- ( - 2 ) \pm 2 3 i}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 3 i}{2 \cdot 1}, u_{2} = \frac{- ( - 2 ) - 2 3 i}{2 \cdot 1}

u = \frac{- ( - 2 ) + 2 3 i}{2 \cdot 1} : 1 + 3 i

\frac{- ( - 2 ) + 2 3 i}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 2 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

2 + 23 i : 2 (1 + 3 i)

2 + 23 i

Rewrite as

= 2 \cdot 1 + 23 i

Factor out common term

2

= 2 (1 + 3 i)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 3 i

u = \frac{- ( - 2 ) - 2 3 i}{2 \cdot 1} : 1 - 3 i

\frac{- ( - 2 ) - 2 3 i}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 2 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

2 - 23 i : 2 (1 - 3 i)

2 - 23 i

Rewrite as

= 2 \cdot 1 - 23 i

Factor out common term

2

= 2 (1 - 3 i)

= \frac{2 ( 1 - 3 i )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 - 3 i

The solutions to the quadratic equation are:

u = 1 + 3 i, u = 1 - 3 i

Substitute back

u = cos (x)

cos (x) = 1 + 3 i, cos (x) = 1 - 3 i

cos (x) = 1 + 3 i, cos (x) = 1 - 3 i

cos (x) = 1 + 3 i :

No Solution

cos (x) = 1 + 3 i

No Solution

cos (x) = 1 - 3 i :

No Solution

cos (x) = 1 - 3 i

No Solution

Combine all the solutions

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

- sin (x) - 2 cot (x) + 5 csc (x) = 0

Remove the ones that don't agree with the equation.

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for -sin(x)-2cot(x)+5csc(x)=0 ?
The general solution for -sin(x)-2cot(x)+5csc(x)=0 is No Solution for x\in\mathbb{R}