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

The general solution for 2sin(2x)=1,0<x<2pi is x= pi/(12),x=(5pi)/(12),x=(13pi)/(12),x=(17pi)/(12)

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

2 sin (2 x) = 1, 0 < x < 2 π

x = \frac{π}{12}, x = \frac{5 π}{12}, x = \frac{13 π}{12}, x = \frac{17 π}{12}

Solution steps

2 sin (2 x) = 1, 0 < x < 2 π

Divide both sides by

2

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

General solutions for

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

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

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solutions for the range

0 < x < 2 π

x = \frac{π}{12}, x = \frac{5 π}{12}, x = \frac{13 π}{12}, x = \frac{17 π}{12}

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

sin (x) = cos (2 x), 0 \leq x \leq 2 π

sin (2 x) = sin (x) cos (x)

sin (θ) = - \frac{5}{5}, \frac{3 π}{2} < θ < 2 π

(2 sin (2 x) - 1) (3 tan (2 x) - 1) = 0

What is the general solution for 2sin(2x)=1,0<x<2pi ?
The general solution for 2sin(2x)=1,0<x<2pi is x= pi/(12),x=(5pi)/(12),x=(13pi)/(12),x=(17pi)/(12)