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

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

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

1 - sin (x) - 2 sin^{2} (x) = 0

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

Solution steps

1 - sin (x) - 2 sin^{2} (x) = 0

Solve by substitution

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

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

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

Combine all the solutions

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

sin^{2} (x) + cos^{2} (x) - 1 + cos (x) - sin (x) = 0

sin (4 x) - cos (2 x) = 0

\frac{6}{π} = 3 cos (2 x)

sin (2 x) = - 2 cos (x)

4 - 8 cos^{2} (x) = 0

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