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

The general solution for (2sin(x)-1)(sin(x)+1)=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

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

Solving each part separately

2 sin (x) - 1 = 0 or sin (x) + 1 = 0

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (A) = \frac{3}{4}

4 sin^{2} (x) + 42 cos (x) - 6 = 0

4 - 5 cos (x) = 4 + 3 sin (x)

\frac{tan ( θ ) cos ( θ )}{csc ( θ )} = sin (θ)

sin (θ) = \frac{44}{12}

What is the general solution for (2sin(x)-1)(sin(x)+1)=0 ?
The general solution for (2sin(x)-1)(sin(x)+1)=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin