What is the general solution for 1+3sin(x)=2 ?

The general solution for 1+3sin(x)=2 is x=0.33983…+2pin,x=pi-0.33983…+2pin

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1+3sin(x)=2

1 + 3 sin (x) = 2

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

Degrees

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

1 + 3 sin (x) = 2

Move

1

to the right side

1 + 3 sin (x) = 2

Subtract

1

from both sides

1 + 3 sin (x) - 1 = 2 - 1

Simplify

3 sin (x) = 1

3 sin (x) = 1

Divide both sides by

3

3 sin (x) = 1

Divide both sides by

3

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

Simplify

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

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

Show solutions in decimal form

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

cos (\frac{x}{2} + \frac{π}{3}) = \frac{1}{2}

sinh (x) = \frac{2}{2}

sin^{2} (2 x) = cos^{2} (x)

- 2 sin (2 x) sin (x) = sin (2 x)

2 sin (θ) - 2 = 0, 0 \leq θ \leq 2 π

What is the general solution for 1+3sin(x)=2 ?
The general solution for 1+3sin(x)=2 is x=0.33983…+2pin,x=pi-0.33983…+2pin