What is the general solution for ((2sin(x)-1))/((sin(x)+5))=0 ?

The general solution for ((2sin(x)-1))/((sin(x)+5))=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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((2sin(x)-1))/((sin(x)+5))=0

\frac{( 2 sin ( x ) - 1 )}{( sin ( x ) + 5 )} = 0

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

Solution steps

\frac{( 2 sin ( x ) - 1 )}{( sin ( x ) + 5 )} = 0

Solve by substitution

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

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

sec^{2} (b) = 2 + tan (b)

cos^{23} (x) + cos^{2} (x) = 0

sin (b) = 0.775

- 2 cos^{2} (x) + 3 sin (x) + 3 = 0

2 cos^{2} (x) = 3 \cdot cos (x)

What is the general solution for ((2sin(x)-1))/((sin(x)+5))=0 ?
The general solution for ((2sin(x)-1))/((sin(x)+5))=0 is x= pi/6+2pin,x=(5pi)/6+2pin