What is the general solution for 4sin(c)+3=sin(c)+1 ?

The general solution for 4sin(c)+3=sin(c)+1 is c=-0.72972…+2pin,c=pi+0.72972…+2pin

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4sin(c)+3=sin(c)+1

4 sin (c) + 3 = sin (c) + 1

c = - 0.72972 \dots + 2 πn, c = π + 0.72972 \dots + 2 πn

Degrees

c = - 41.81031 \dots^{\circ} + 36 0^{\circ} n, c = 221.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (c) + 3 = sin (c) + 1

Solve by substitution

4 sin (c) + 3 = sin (c) + 1

Let:

sin (c) = u

4 u + 3 = u + 1

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

4 u + 3 = u + 1

Move

3

to the right side

4 u + 3 = u + 1

Subtract

3

from both sides

4 u + 3 - 3 = u + 1 - 3

Simplify

4 u = u - 2

4 u = u - 2

Move

u

to the left side

4 u = u - 2

Subtract

u

from both sides

4 u - u = u - 2 - u

Simplify

3 u = - 2

3 u = - 2

Divide both sides by

3

3 u = - 2

Divide both sides by

3

\frac{3 u}{3} = \frac{- 2}{3}

Simplify

u = - \frac{2}{3}

u = - \frac{2}{3}

Substitute back

u = sin (c)

sin (c) = - \frac{2}{3}

sin (c) = - \frac{2}{3}

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

sin (c) = - \frac{2}{3}

Apply trig inverse properties

sin (c) = - \frac{2}{3}

General solutions for

sin (c) = - \frac{2}{3}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

c = - 0.72972 \dots + 2 πn, c = π + 0.72972 \dots + 2 πn

(tan (x) + 1) (3 tan (x) - 1) = 0

tan^{2} (x) sin (x) - sin (x) = 0

2 cos^{2} (t) - 3 cos (t) + 1 = 0

sin (x + 3 0^{\circ}) = cos (x)

sin (\frac{π}{6} x) = 0

What is the general solution for 4sin(c)+3=sin(c)+1 ?
The general solution for 4sin(c)+3=sin(c)+1 is c=-0.72972…+2pin,c=pi+0.72972…+2pin