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

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

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sin(x)+2=-csc(x)

Solution

sin (x) + 2 = - csc (x)

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

sin (x) + 2 = - csc (x)

Subtract

- csc (x)

from both sides

sin (x) + 2 + csc (x) = 0

Rewrite using trig identities

2 + csc (x) + sin (x)

Use the basic trigonometric identity:

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

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

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

Solve by substitution

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

Let:

csc (x) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

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

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

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

Solve

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 2, c = 1

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

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

2^{2} - 4 \cdot 1 \cdot 1 = 0

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

2 + u + \frac{1}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1

Substitute back

u = csc (x)

csc (x) = - 1

csc (x) = - 1

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

Popular Examples

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cos (a) = \frac{9}{14}

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\frac{π}{2} = 2 sin (x)

cos(x)=(sin(x))^2

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

(20)/(sin(35))=(16)/(sin(C))

\frac{20}{sin ( 3 5 ^{\circ} )} = \frac{16}{sin ( C )}

(tan(2θ))/2 =tan(θ)

\frac{tan ( 2 θ )}{2} = tan (θ)

Frequently Asked Questions (FAQ)

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