What is the general solution for 6/pi-4sin(2x)=0 ?

The general solution for 6/pi-4sin(2x)=0 is x=(0.49776…)/2+pin,x= pi/2-(0.49776…)/2+pin

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6/pi-4sin(2x)=0

\frac{6}{π} - 4 sin (2 x) = 0

x = \frac{0.49776 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.49776 \dots}{2} + πn

Solution steps

\frac{6}{π} - 4 sin (2 x) = 0

Move

\frac{6}{π}

to the right side

- 4 sin (2 x) = - \frac{6}{π}

Divide both sides by

- 4

sin (2 x) = \frac{3}{2 π}

Apply trig inverse properties

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

Solve

2 x = arcsin (\frac{3}{2 π}) + 2 πn : x = \frac{arcsin ( \frac{3}{2 π} )}{2} + πn

Solve

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

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

Show solutions in decimal form

x = \frac{0.49776 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.49776 \dots}{2} + πn

sin (t) = \frac{1}{5}

sin^{2} (x) + \frac{1}{2} sin (x) = 0

sin (θ) = \frac{- 1}{2}

(sin (a) + cos (a))^{2} = 1 + 2, a cos (a)

4 sin (z) cos^{2} (z) = 3 sin (z)

What is the general solution for 6/pi-4sin(2x)=0 ?
The general solution for 6/pi-4sin(2x)=0 is x=(0.49776…)/2+pin,x= pi/2-(0.49776…)/2+pin