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=(7pi)/6+2pin,x=(11pi)/6+2pin,x= pi/2+2pin

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

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

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

Solution steps

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

Solve by substitution

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

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

Combine all the solutions

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

y, z = e^{- x} sin (x + 2 y)

3 sin (θ) - 2 sin^{2} (θ) = 0

3 tan (\frac{x}{3}) + 3 = 0

0 = 2 sin (2 x)

sec^{2} (x) = tan (x) + 1

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=(7pi)/6+2pin,x=(11pi)/6+2pin,x= pi/2+2pin