What is the general solution for 2*sin^2(x)-sin(x)=1 ?

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

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2*sin^2(x)-sin(x)=1

2 \cdot sin^{2} (x) - sin (x) = 1

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

Solution steps

2 sin^{2} (x) - sin (x) = 1

Solve by substitution

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

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

4 cos^{2} (3 x) - 1 = 0

6 sin (b) + 12 = 0

40 + 40 cos (\frac{2 π x}{9}) = 50

tan (x) + sec (x) - 1 = 0

sin (x) (sin (x) - 1) = 3 cos^{2} (x)

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