What is the general solution for 8sin(θ)cos(θ)tan(θ)=csc(θ) ?

The general solution for 8sin(θ)cos(θ)tan(θ)=csc(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin

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8sin(θ)cos(θ)tan(θ)=csc(θ)

Solution

8 sin (θ) cos (θ) tan (θ) = csc (θ)

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

8 sin (θ) cos (θ) tan (θ) = csc (θ)

Subtract

csc (θ)

from both sides

8 sin (θ) cos (θ) tan (θ) - csc (θ) = 0

Express with sin, cos

- csc (θ) + 8 cos (θ) sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= - \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= - \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

- \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )} : \frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )}

- \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )} = 8 sin^{2} (θ)

8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) \cdot 8 cos ( θ ) sin ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= sin (θ) \cdot 8 sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 8 sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 8 sin^{2} (θ)

= - \frac{1}{sin ( θ )} + 8 sin^{2} (θ)

Convert element to fraction:

8 sin^{2} (θ) = \frac{8 sin ^{2} ( θ ) sin ( θ )}{sin ( θ )}

= - \frac{1}{sin ( θ )} + \frac{8 sin ^{2} ( θ ) sin ( θ )}{sin ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 + 8 sin ^{2} ( θ ) sin ( θ )}{sin ( θ )}

- 1 + 8 sin^{2} (θ) sin (θ) = - 1 + 8 sin^{3} (θ)

- 1 + 8 sin^{2} (θ) sin (θ)

8 sin^{2} (θ) sin (θ) = 8 sin^{3} (θ)

8 sin^{2} (θ) sin (θ)

Apply exponent rule:

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

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= 8 sin^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 8 sin^{3} (θ)

= - 1 + 8 sin^{3} (θ)

= \frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )}

= \frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )}

\frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )} = 0

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

- 1 + 8 sin^{3} (θ) = 0

Solve by substitution

- 1 + 8 sin^{3} (θ) = 0

Let:

sin (θ) = u

- 1 + 8 u^{3} = 0

- 1 + 8 u^{3} = 0 : u = \frac{1}{2}, u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

- 1 + 8 u^{3} = 0

Move

1

to the right side

- 1 + 8 u^{3} = 0

Add

1

to both sides

- 1 + 8 u^{3} + 1 = 0 + 1

Simplify

8 u^{3} = 1

8 u^{3} = 1

Divide both sides by

8

8 u^{3} = 1

Divide both sides by

8

\frac{8 u ^{3}}{8} = \frac{1}{8}

Simplify

u^{3} = \frac{1}{8}

u^{3} = \frac{1}{8}

For

x^{3} = f (a)

the solutions are

Apply radical rule:

assuming

a \geq 0, b \geq 0

Factor the number:

8 = 2^{3}

Apply radical rule:

= 2

Apply rule

= \frac{1}{2}

Simplify

Apply radical rule:

assuming

a \geq 0, b \geq 0

Factor the number:

8 = 2^{3}

Apply radical rule:

= 2

Apply rule

= \frac{1}{2}

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

Multiply fractions:

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

= \frac{1 \cdot ( - 1 + 3 i )}{2 \cdot 2}

1 \cdot (- 1 + 3 i) = - 1 + 3 i

1 \cdot (- 1 + 3 i)

Multiply:

1 \cdot (- 1 + 3 i) = (- 1 + 3 i)

= (- 1 + 3 i)

Remove parentheses:

(- a) = - a

= - 1 + 3 i

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 3 i}{4}

Rewrite

\frac{- 1 + 3 i}{4}

in standard complex form:

- \frac{1}{4} + \frac{3}{4} i

\frac{- 1 + 3 i}{4}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{4} = - \frac{1}{4} + \frac{3 i}{4}

= - \frac{1}{4} + \frac{3 i}{4}

= - \frac{1}{4} + \frac{3}{4} i

Simplify

Apply radical rule:

assuming

a \geq 0, b \geq 0

Factor the number:

8 = 2^{3}

Apply radical rule:

= 2

Apply rule

= \frac{1}{2}

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

Multiply fractions:

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

= \frac{1 \cdot ( - 1 - 3 i )}{2 \cdot 2}

1 \cdot (- 1 - 3 i) = - 1 - 3 i

1 \cdot (- 1 - 3 i)

Multiply:

1 \cdot (- 1 - 3 i) = (- 1 - 3 i)

= (- 1 - 3 i)

Remove parentheses:

(- a) = - a

= - 1 - 3 i

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 3 i}{4}

Rewrite

\frac{- 1 - 3 i}{4}

in standard complex form:

- \frac{1}{4} - \frac{3}{4} i

\frac{- 1 - 3 i}{4}

Apply the fraction rule:

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

\frac{- 1 - 3 i}{4} = - \frac{1}{4} - \frac{3 i}{4}

= - \frac{1}{4} - \frac{3 i}{4}

= - \frac{1}{4} - \frac{3}{4} i

u = \frac{1}{2}, u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{4} + i \frac{3}{4}, sin (θ) = - \frac{1}{4} - i \frac{3}{4}

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{4} + i \frac{3}{4}, sin (θ) = - \frac{1}{4} - i \frac{3}{4}

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

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

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - \frac{1}{4} + i \frac{3}{4} :

No Solution

sin (θ) = - \frac{1}{4} + i \frac{3}{4}

No Solution

sin (θ) = - \frac{1}{4} - i \frac{3}{4} :

No Solution

sin (θ) = - \frac{1}{4} - i \frac{3}{4}

No Solution

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 8sin(θ)cos(θ)tan(θ)=csc(θ) ?
The general solution for 8sin(θ)cos(θ)tan(θ)=csc(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin