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

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

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2=2sin(x)

2 = 2 sin (x)

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

2 = 2 sin (x)

Switch sides

2 sin (x) = 2

Divide both sides by

2

2 sin (x) = 2

Divide both sides by

2

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

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

cos^{2} (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

cot (θ) - 3 = 0, 0 \leq θ \leq 2 π

2 = sec (θ)

sin (x) = \frac{1}{8}, sin (2 x)

x, csc (x) = 2

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