What is the general solution for sin(x)-csc(x)+1=0 ?

The general solution for sin(x)-csc(x)+1=0 is x=0.66623…+2pin,x=pi-0.66623…+2pin

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sin(x)-csc(x)+1=0

Solution

sin (x) - csc (x) + 1 = 0

Solution

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

Degrees

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) - csc (x) + 1 = 0

Rewrite using trig identities

1 - csc (x) + sin (x)

Use the basic trigonometric identity:

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

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

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

Solve by substitution

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

Let:

csc (x) = u

1 - u + \frac{1}{u} = 0

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

1 - u + \frac{1}{u} = 0

Multiply both sides by

u

1 - u + \frac{1}{u} = 0

Multiply both sides by

u

1 \cdot u - uu + \frac{1}{u} u = 0 \cdot u

Simplify

1 \cdot u - uu + \frac{1}{u} u = 0 \cdot u

Simplify

1 \cdot u : u

1 \cdot u

Multiply:

1 \cdot u = u

= u

Simplify

- uu : - u^{2}

- uu

Apply exponent rule:

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

uu = u^{1 + 1}

= - u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - u^{2}

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

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u - u^{2} + 1 = 0

u - u^{2} + 1 = 0

u - u^{2} + 1 = 0

Solve

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

u - u^{2} + 1 = 0

Write in the standard form

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

- u^{2} + u + 1 = 0

Solve with the quadratic formula

- u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 1

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

1 - u + \frac{1}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

Substitute back

u = csc (x)

csc (x) = - \frac{- 1 + 5}{2}, csc (x) = \frac{1 + 5}{2}

csc (x) = - \frac{- 1 + 5}{2}, csc (x) = \frac{1 + 5}{2}

csc (x) = - \frac{- 1 + 5}{2} :

No Solution

csc (x) = - \frac{- 1 + 5}{2}

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

No Solution

csc (x) = \frac{1 + 5}{2} : x = arccsc (\frac{1 + 5}{2}) + 2 πn, x = π - arccsc (\frac{1 + 5}{2}) + 2 πn

csc (x) = \frac{1 + 5}{2}

Apply trig inverse properties

csc (x) = \frac{1 + 5}{2}

General solutions for

csc (x) = \frac{1 + 5}{2}

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (\frac{1 + 5}{2}) + 2 πn, x = π - arccsc (\frac{1 + 5}{2}) + 2 πn

x = arccsc (\frac{1 + 5}{2}) + 2 πn, x = π - arccsc (\frac{1 + 5}{2}) + 2 πn

Combine all the solutions

x = arccsc (\frac{1 + 5}{2}) + 2 πn, x = π - arccsc (\frac{1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)-csc(x)+1=0 ?
The general solution for sin(x)-csc(x)+1=0 is x=0.66623…+2pin,x=pi-0.66623…+2pin