What is the general solution for sin(x)=1,2pi<= x<= 4pi ?

The general solution for sin(x)=1,2pi<= x<= 4pi is x=(5pi)/2

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sin(x)=1,2pi<= x<= 4pi

sin (x) = 1, 2 π \leq x \leq 4 π

x = \frac{5 π}{2}

Degrees

x = 45 0^{\circ}

Solution steps

sin (x) = 1, 2 π \leq x \leq 4 π

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

Solutions for the range

2 π \leq x \leq 4 π

x = \frac{5 π}{2}

cos (a) = 1

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

sec (θ) = \frac{7}{5}

\frac{sin ( x )}{cos ( x )} = 0

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

What is the general solution for sin(x)=1,2pi<= x<= 4pi ?
The general solution for sin(x)=1,2pi<= x<= 4pi is x=(5pi)/2