What is the general solution for -sin(B)-4=3sin(B)-5 ?

The general solution for -sin(B)-4=3sin(B)-5 is B=0.25268…+2pin,B=pi-0.25268…+2pin

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-sin(B)-4=3sin(B)-5

- sin (B) - 4 = 3 sin (B) - 5

B = 0.25268 \dots + 2 πn, B = π - 0.25268 \dots + 2 πn

Degrees

B = 14.47751 \dots^{\circ} + 36 0^{\circ} n, B = 165.52248 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- sin (B) - 4 = 3 sin (B) - 5

Solve by substitution

- sin (B) - 4 = 3 sin (B) - 5

Let:

sin (B) = u

- u - 4 = 3 u - 5

- u - 4 = 3 u - 5 : u = \frac{1}{4}

- u - 4 = 3 u - 5

Move

4

to the right side

- u - 4 = 3 u - 5

Add

4

to both sides

- u - 4 + 4 = 3 u - 5 + 4

Simplify

- u = 3 u - 1

- u = 3 u - 1

Move

3 u

to the left side

- u = 3 u - 1

Subtract

3 u

from both sides

- u - 3 u = 3 u - 1 - 3 u

Simplify

- 4 u = - 1

- 4 u = - 1

Divide both sides by

- 4

- 4 u = - 1

Divide both sides by

- 4

\frac{- 4 u}{- 4} = \frac{- 1}{- 4}

Simplify

u = \frac{1}{4}

u = \frac{1}{4}

Substitute back

u = sin (B)

sin (B) = \frac{1}{4}

sin (B) = \frac{1}{4}

sin (B) = \frac{1}{4} : B = arcsin (\frac{1}{4}) + 2 πn, B = π - arcsin (\frac{1}{4}) + 2 πn

sin (B) = \frac{1}{4}

Apply trig inverse properties

sin (B) = \frac{1}{4}

General solutions for

sin (B) = \frac{1}{4}

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

B = arcsin (\frac{1}{4}) + 2 πn, B = π - arcsin (\frac{1}{4}) + 2 πn

B = arcsin (\frac{1}{4}) + 2 πn, B = π - arcsin (\frac{1}{4}) + 2 πn

Combine all the solutions

B = arcsin (\frac{1}{4}) + 2 πn, B = π - arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

B = 0.25268 \dots + 2 πn, B = π - 0.25268 \dots + 2 πn

cos^{2} (x) - 7 cos (x) - 1 = 0

sec^{2} (x) + tan (x) - 1 = 0

0 = sin (θ)

6 tan^{4} (x) - tan^{2} (x) - 2 = 0

t, x = sin (t)

What is the general solution for -sin(B)-4=3sin(B)-5 ?
The general solution for -sin(B)-4=3sin(B)-5 is B=0.25268…+2pin,B=pi-0.25268…+2pin