What is the general solution for solvefor θ,2sin(θ)=csc(θ)+1 ?

The general solution for solvefor θ,2sin(θ)=csc(θ)+1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin

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Popular >

solvefor θ,2sin(θ)=csc(θ)+1

Solution

solve for

θ, 2 sin (θ) = csc (θ) + 1

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) = csc (θ) + 1

Subtract

csc (θ) + 1

from both sides

2 sin (θ) - csc (θ) - 1 = 0

Rewrite using trig identities

- 1 - csc (θ) + 2 sin (θ)

Use the basic trigonometric identity:

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

= - 1 - csc (θ) + 2 \cdot \frac{1}{csc ( θ )}

2 \cdot \frac{1}{csc ( θ )} = \frac{2}{csc ( θ )}

2 \cdot \frac{1}{csc ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 2}{csc ( θ )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{csc ( θ )}

= - 1 - csc (θ) + \frac{2}{csc ( θ )}

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

Solve by substitution

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

Let:

csc (θ) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

- 1 \cdot u - uu + \frac{2}{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{2}{u} u : 2

\frac{2}{u} u

Multiply fractions:

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

= \frac{2 u}{u}

Cancel the common factor:

u

= 2

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 1) \cdot 2 = 3

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

1 + 3 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{- 2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

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

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

Remove parentheses:

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

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

Subtract the numbers:

1 - 3 = - 2

= \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

The solutions to the quadratic equation are:

u = - 2, u = 1

u = - 2, u = 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 2, u = 1

Substitute back

u = csc (θ)

csc (θ) = - 2, csc (θ) = 1

csc (θ) = - 2, csc (θ) = 1

csc (θ) = - 2 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

csc (θ) = - 2

General solutions for

csc (θ) = - 2

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

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

csc (θ) = 1 : θ = \frac{π}{2} + 2 πn

csc (θ) = 1

General solutions for

csc (θ) = 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

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor θ,2sin(θ)=csc(θ)+1 ?
The general solution for solvefor θ,2sin(θ)=csc(θ)+1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin