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

The general solution for (sin(x)-1)(2sin(x)+1)=0 is x= pi/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin

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

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

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solution steps

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

Solving each part separately

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

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

2 sin (x) + 1 = 0 : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

0 = arctan (x)

2 tan (x) + 3 = 5

cos (a) = \frac{3}{2}

tan^{2} (x) + tan (x) - 12 = 0, 0 \leq x \leq 2 π

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

What is the general solution for (sin(x)-1)(2sin(x)+1)=0 ?
The general solution for (sin(x)-1)(2sin(x)+1)=0 is x= pi/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin