What is the general solution for 2sin(x)=1,0<= x<= 2pi ?

The general solution for 2sin(x)=1,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6

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2sin(x)=1,0<= x<= 2pi

2 sin (x) = 1, 0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Degrees

x = 3 0^{\circ}, x = 15 0^{\circ}

Solution steps

2 sin (x) = 1, 0 \leq x \leq 2 π

Divide both sides by

2

2 sin (x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

5 cot (x) + 4 = 9

sin^{5} (x) = 0

3 sin (4 x) + 5 = 6, 0 \leq x \leq \frac{π}{2}

cos (z + 1) = 0

4 csc^{2} (\frac{x}{2}) - 6 csc (\frac{x}{2}) - 4 = 0

What is the general solution for 2sin(x)=1,0<= x<= 2pi ?
The general solution for 2sin(x)=1,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6