What is the general solution for 5sin(a)+3=sin(a)+2 ?

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

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5sin(a)+3=sin(a)+2

5 sin (a) + 3 = sin (a) + 2

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

Degrees

a = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, a = 194.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (a) + 3 = sin (a) + 2

Solve by substitution

5 sin (a) + 3 = sin (a) + 2

Let:

sin (a) = u

5 u + 3 = u + 2

5 u + 3 = u + 2 : u = - \frac{1}{4}

5 u + 3 = u + 2

Move

3

to the right side

5 u + 3 = u + 2

Subtract

3

from both sides

5 u + 3 - 3 = u + 2 - 3

Simplify

5 u = u - 1

5 u = u - 1

Move

u

to the left side

5 u = u - 1

Subtract

u

from both sides

5 u - u = u - 1 - 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 (a)

sin (a) = - \frac{1}{4}

sin (a) = - \frac{1}{4}

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

sin (a) = - \frac{1}{4}

Apply trig inverse properties

sin (a) = - \frac{1}{4}

General solutions for

sin (a) = - \frac{1}{4}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

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

csc (3 x) = 2

sin (x) = 0, 6

4 cos^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

2 sin (θ) - 3 = - csc (θ)

cot (θ) = - 3

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