What is the general solution for 4sin(x)+5=4+sin(x) ?

The general solution for 4sin(x)+5=4+sin(x) is x=-0.33983…+2pin,x=pi+0.33983…+2pin

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4sin(x)+5=4+sin(x)

4 sin (x) + 5 = 4 + sin (x)

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

Degrees

x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (x) + 5 = 4 + sin (x)

Solve by substitution

4 sin (x) + 5 = 4 + sin (x)

Let:

sin (x) = u

4 u + 5 = 4 + u

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

4 u + 5 = 4 + u

Move

5

to the right side

4 u + 5 = 4 + u

Subtract

5

from both sides

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

Simplify

4 u = u - 1

4 u = u - 1

Move

u

to the left side

4 u = u - 1

Subtract

u

from both sides

4 u - u = u - 1 - u

Simplify

3 u = - 1

3 u = - 1

Divide both sides by

3

3 u = - 1

Divide both sides by

3

\frac{3 u}{3} = \frac{- 1}{3}

Simplify

u = - \frac{1}{3}

u = - \frac{1}{3}

Substitute back

u = sin (x)

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

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

sin (x) = - \frac{1}{3} : x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

Show solutions in decimal form

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

9 sin (θ) - 1 = 7 sin (θ)

cot (x) = sin (x)

- 3 cos (x) - sin (x) = - 3

9 cot^{2} (x) = 3

cos (x) = - 2

What is the general solution for 4sin(x)+5=4+sin(x) ?
The general solution for 4sin(x)+5=4+sin(x) is x=-0.33983…+2pin,x=pi+0.33983…+2pin