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

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

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3sin(x)-1=2

3 sin (x) - 1 = 2

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) - 1 = 2

Move

1

to the right side

3 sin (x) - 1 = 2

Add

1

to both sides

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

Simplify

3 sin (x) = 3

3 sin (x) = 3

Divide both sides by

3

3 sin (x) = 3

Divide both sides by

3

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

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (t) = \frac{1}{2}

- 2 sin^{2} (θ) + 5 sin (θ) = 2

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

3 sin (x) - 2 = 0

4 cos^{2} (x) - 2 sin (x) + 1 = 0

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