What is the general solution for (sin(2x))/(4sin(2x)-1)=1 ?

The general solution for (sin(2x))/(4sin(2x)-1)=1 is x=(0.33983…)/2+pin,x= pi/2-(0.33983…)/2+pin

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(sin(2x))/(4sin(2x)-1)=1

\frac{sin ( 2 x )}{4 sin ( 2 x ) - 1} = 1

x = \frac{0.33983 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

Solution steps

\frac{sin ( 2 x )}{4 sin ( 2 x ) - 1} = 1

Solve by substitution

sin (2 x) = \frac{1}{3}

sin (2 x) = \frac{1}{3} : x = \frac{arcsin ( \frac{1}{3} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

Combine all the solutions

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

Show solutions in decimal form

x = \frac{0.33983 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

arctan (x) = 119.74

2 cos (θ) sin (θ) = sin (θ)

4 cos (x) = sec (x)

9.6 = 11.6 cos (3.922 t)

2 sin^{2} (x) = tan^{2} (x)

What is the general solution for (sin(2x))/(4sin(2x)-1)=1 ?
The general solution for (sin(2x))/(4sin(2x)-1)=1 is x=(0.33983…)/2+pin,x= pi/2-(0.33983…)/2+pin