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Popular >

log_{8}(4)=((sin^2(x)))/(sin(x)-1/8)

Solution

lo g_{8} (4) = \frac{( sin ^{2} ( x ) )}{sin ( x ) - \frac{1}{8}}

Solution

x = 0.16744 \dots + 2 πn, x = π - 0.16744 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 9.59406 \dots^{\circ} + 36 0^{\circ} n, x = 170.40593 \dots^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

lo g_{8} (4) = \frac{( sin ^{2} ( x ) )}{sin ( x ) - \frac{1}{8}}

Subtract

\frac{sin ^{2} ( x )}{sin ( x ) - \frac{1}{8}}

from both sides

\frac{2}{3} - \frac{8 sin ^{2} ( x )}{8 sin ( x ) - 1} = 0

Simplify

\frac{2}{3} - \frac{8 sin ^{2} ( x )}{8 sin ( x ) - 1} : \frac{2 ( 8 sin ( x ) - 1 ) - 24 sin ^{2} ( x )}{3 ( 8 sin ( x ) - 1 )}

\frac{2}{3} - \frac{8 sin ^{2} ( x )}{8 sin ( x ) - 1}

Least Common Multiplier of

3, 8 sin (x) - 1 : 3 (8 sin (x) - 1)

3, 8 sin (x) - 1

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

3

8 sin (x) - 1

= 3 (8 sin (x) - 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

3 (8 sin (x) - 1)

For

\frac{2}{3} :

multiply the denominator and numerator by

8 sin (x) - 1

\frac{2}{3} = \frac{2 ( 8 sin ( x ) - 1 )}{3 ( 8 sin ( x ) - 1 )}

For

\frac{8 sin ^{2} ( x )}{8 sin ( x ) - 1} :

multiply the denominator and numerator by

3

\frac{8 sin ^{2} ( x )}{8 sin ( x ) - 1} = \frac{8 sin ^{2} ( x ) \cdot 3}{( 8 sin ( x ) - 1 ) \cdot 3} = \frac{24 sin ^{2} ( x )}{3 ( 8 sin ( x ) - 1 )}

= \frac{2 ( 8 sin ( x ) - 1 )}{3 ( 8 sin ( x ) - 1 )} - \frac{24 sin ^{2} ( x )}{3 ( 8 sin ( x ) - 1 )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 ( 8 sin ( x ) - 1 ) - 24 sin ^{2} ( x )}{3 ( 8 sin ( x ) - 1 )}

\frac{2 ( 8 sin ( x ) - 1 ) - 24 sin ^{2} ( x )}{3 ( 8 sin ( x ) - 1 )} = 0

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

2 (8 sin (x) - 1) - 24 sin^{2} (x) = 0

Solve by substitution

2 (8 sin (x) - 1) - 24 sin^{2} (x) = 0

Let:

sin (x) = u

2 (8 u - 1) - 24 u^{2} = 0

2 (8 u - 1) - 24 u^{2} = 0 : u = \frac{1}{6}, u = \frac{1}{2}

2 (8 u - 1) - 24 u^{2} = 0

Expand

2 (8 u - 1) - 24 u^{2} : 16 u - 2 - 24 u^{2}

2 (8 u - 1) - 24 u^{2}

Expand

2 (8 u - 1) : 16 u - 2

2 (8 u - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 8 u, c = 1

= 2 \cdot 8 u - 2 \cdot 1

Simplify

2 \cdot 8 u - 2 \cdot 1 : 16 u - 2

2 \cdot 8 u - 2 \cdot 1

Multiply the numbers:

2 \cdot 8 = 16

= 16 u - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 16 u - 2

= 16 u - 2

= 16 u - 2 - 24 u^{2}

16 u - 2 - 24 u^{2} = 0

Write in the standard form

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

- 24 u^{2} + 16 u - 2 = 0

Solve with the quadratic formula

- 24 u^{2} + 16 u - 2 = 0

Quadratic Equation Formula:

For

a = - 24, b = 16, c = - 2

u_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 24 ) ( - 2 )}{2 ( - 24 )}

u_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 24 ) ( - 2 )}{2 ( - 24 )}

1 6^{2} - 4 (- 24) (- 2) = 8

1 6^{2} - 4 (- 24) (- 2)

Apply rule

- (- a) = a

= 1 6^{2} - 4 \cdot 24 \cdot 2

Multiply the numbers:

4 \cdot 24 \cdot 2 = 192

= 1 6^{2} - 192

1 6^{2} = 256

= 256 - 192

Subtract the numbers:

256 - 192 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 16 \pm 8}{2 ( - 24 )}

Separate the solutions

u_{1} = \frac{- 16 + 8}{2 ( - 24 )}, u_{2} = \frac{- 16 - 8}{2 ( - 24 )}

u = \frac{- 16 + 8}{2 ( - 24 )} : \frac{1}{6}

\frac{- 16 + 8}{2 ( - 24 )}

Remove parentheses:

(- a) = - a

= \frac{- 16 + 8}{- 2 \cdot 24}

Add/Subtract the numbers:

- 16 + 8 = - 8

= \frac{- 8}{- 2 \cdot 24}

Multiply the numbers:

2 \cdot 24 = 48

= \frac{- 8}{- 48}

Apply the fraction rule:

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

= \frac{8}{48}

Cancel the common factor:

8

= \frac{1}{6}

u = \frac{- 16 - 8}{2 ( - 24 )} : \frac{1}{2}

\frac{- 16 - 8}{2 ( - 24 )}

Remove parentheses:

(- a) = - a

= \frac{- 16 - 8}{- 2 \cdot 24}

Subtract the numbers:

- 16 - 8 = - 24

= \frac{- 24}{- 2 \cdot 24}

Multiply the numbers:

2 \cdot 24 = 48

= \frac{- 24}{- 48}

Apply the fraction rule:

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

= \frac{24}{48}

Cancel the common factor:

24

= \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{1}{6}, u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = \frac{1}{6}, sin (x) = \frac{1}{2}

sin (x) = \frac{1}{6}, sin (x) = \frac{1}{2}

sin (x) = \frac{1}{6} : x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

sin (x) = \frac{1}{6}

Apply trig inverse properties

sin (x) = \frac{1}{6}

General solutions for

sin (x) = \frac{1}{6}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

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

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

General solutions for

sin (x) = \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}

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

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

Combine all the solutions

x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Show solutions in decimal form

x = 0.16744 \dots + 2 πn, x = π - 0.16744 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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