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

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

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

sin (x) - 1 = 0, 0 \leq x < 2 π

x = \frac{π}{2}

Solution steps

sin (x) - 1 = 0, 0 \leq x < 2 π

Move

1

to the right side

sin (x) = 1

General solutions for

sin (x) = 1

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}

3 cos (θ) = 2 cos (θ)

\frac{26.4575}{sin ( 6 0 ^{\circ} )} = \frac{20}{sin ( B )}

sin (\frac{5}{2} θ) = - 0.966

sin^{2} (x) - \frac{2}{3} = 0

0 = - 8 sin (x)

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