What is the general solution for 4csc(x)+sin(x)+5=0 ?

The general solution for 4csc(x)+sin(x)+5=0 is x=(3pi)/2+2pin

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4csc(x)+sin(x)+5=0

Solution

4 csc (x) + sin (x) + 5 = 0

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

4 csc (x) + sin (x) + 5 = 0

Rewrite using trig identities

5 + sin (x) + 4 csc (x)

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

= 5 + \frac{1}{csc ( x )} + 4 csc (x)

5 + \frac{1}{csc ( x )} + 4 csc (x) = 0

Solve by substitution

5 + \frac{1}{csc ( x )} + 4 csc (x) = 0

Let:

csc (x) = u

5 + \frac{1}{u} + 4 u = 0

5 + \frac{1}{u} + 4 u = 0 : u = - \frac{1}{4}, u = - 1

5 + \frac{1}{u} + 4 u = 0

Multiply both sides by

u

5 + \frac{1}{u} + 4 u = 0

Multiply both sides by

u

5 u + \frac{1}{u} u + 4 uu = 0 \cdot u

Simplify

5 u + \frac{1}{u} u + 4 uu = 0 \cdot u

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

4 uu : 4 u^{2}

4 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 4 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

5 u + 1 + 4 u^{2} = 0

5 u + 1 + 4 u^{2} = 0

5 u + 1 + 4 u^{2} = 0

Solve

5 u + 1 + 4 u^{2} = 0 : u = - \frac{1}{4}, u = - 1

5 u + 1 + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} + 5 u + 1 = 0

Solve with the quadratic formula

4 u^{2} + 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 4, b = 5, c = 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

5^{2} - 4 \cdot 4 \cdot 1 = 3

5^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 5 + 3}{2 \cdot 4}, u_{2} = \frac{- 5 - 3}{2 \cdot 4}

u = \frac{- 5 + 3}{2 \cdot 4} : - \frac{1}{4}

\frac{- 5 + 3}{2 \cdot 4}

Add/Subtract the numbers:

- 5 + 3 = - 2

= \frac{- 2}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

u = \frac{- 5 - 3}{2 \cdot 4} : - 1

\frac{- 5 - 3}{2 \cdot 4}

Subtract the numbers:

- 5 - 3 = - 8

= \frac{- 8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - \frac{1}{4}, u = - 1

u = - \frac{1}{4}, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

5 + \frac{1}{u} + 4 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - \frac{1}{4}, u = - 1

Substitute back

u = csc (x)

csc (x) = - \frac{1}{4}, csc (x) = - 1

csc (x) = - \frac{1}{4}, csc (x) = - 1

csc (x) = - \frac{1}{4} :

No Solution

csc (x) = - \frac{1}{4}

csc (x) \leq - 1 or csc (x) \geq 1

No Solution

csc (x) = - 1 : x = \frac{3 π}{2} + 2 πn

csc (x) = - 1

General solutions for

csc (x) = - 1

csc (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4csc(x)+sin(x)+5=0 ?
The general solution for 4csc(x)+sin(x)+5=0 is x=(3pi)/2+2pin