What is the general solution for 2csc(θ)=sqrt(2)cot(θ)+3csc(θ) ?

The general solution for 2csc(θ)=sqrt(2)cot(θ)+3csc(θ) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin

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2csc(θ)=sqrt(2)cot(θ)+3csc(θ)

Solution

2 csc (θ) = 2 cot (θ) + 3 csc (θ)

Solution

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Degrees

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

2 csc (θ) = 2 cot (θ) + 3 csc (θ)

Subtract

2 cot (θ) + 3 csc (θ)

from both sides

- csc (θ) - 2 cot (θ) = 0

Express with sin, cos

- csc (θ) - cot (θ) 2

Use the basic trigonometric identity:

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

= - \frac{1}{sin ( θ )} - cot (θ) 2

Use the basic trigonometric identity:

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

= - \frac{1}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )} 2

Simplify

- \frac{1}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )} 2 : \frac{- 1 - 2 cos ( θ )}{sin ( θ )}

- \frac{1}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )} 2

Multiply

\frac{cos ( θ )}{sin ( θ )} 2 : \frac{2 cos ( θ )}{sin ( θ )}

\frac{cos ( θ )}{sin ( θ )} 2

Multiply fractions:

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

= \frac{cos ( θ ) 2}{sin ( θ )}

= - \frac{1}{sin ( θ )} - \frac{2 cos ( θ )}{sin ( θ )}

Apply rule

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

= \frac{- 1 - 2 cos ( θ )}{sin ( θ )}

= \frac{- 1 - 2 cos ( θ )}{sin ( θ )}

\frac{- 1 - cos ( θ ) 2}{sin ( θ )} = 0

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

- 1 - cos (θ) 2 = 0

Move

1

to the right side

- 1 - cos (θ) 2 = 0

Add

1

to both sides

- 1 - cos (θ) 2 + 1 = 0 + 1

Simplify

- cos (θ) 2 = 1

- cos (θ) 2 = 1

Divide both sides by

- 2

- cos (θ) 2 = 1

Divide both sides by

- 2

\frac{- cos ( θ ) 2}{- 2} = \frac{1}{- 2}

Simplify

\frac{- cos ( θ ) 2}{- 2} = \frac{1}{- 2}

Simplify

\frac{- cos ( θ ) 2}{- 2} : cos (θ)

\frac{- cos ( θ ) 2}{- 2}

Apply the fraction rule:

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

= \frac{cos ( θ ) 2}{2}

Cancel the common factor:

2

= cos (θ)

Simplify

\frac{1}{- 2} : - \frac{2}{2}

\frac{1}{- 2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Popular Examples

sin(1/2 x)=-1

sin (\frac{1}{2} x) = - 1

9cos^2(x)tan(x)=4tan(x)

9 cos^{2} (x) tan (x) = 4 tan (x)

3tan(x)-3=0,0<= x<= 2pi

3 tan (x) - 3 = 0, 0 \leq x \leq 2 π

(sin(57.22))/(31.5)=(sin(Q))/(24.4)

\frac{sin ( 57.2 2 ^{\circ} )}{31.5} = \frac{sin ( Q )}{24.4}

cosh(x)=10

cosh (x) = 10

Frequently Asked Questions (FAQ)

What is the general solution for 2csc(θ)=sqrt(2)cot(θ)+3csc(θ) ?
The general solution for 2csc(θ)=sqrt(2)cot(θ)+3csc(θ) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin