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

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

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1+sin^2(x)=2

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

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

Solution steps

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

Solve by substitution

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

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

Combine all the solutions

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

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

cos (x) tan (x) = 0

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

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

2 sin^{2} (x) - sin (x) - 1 = 0, 0 \leq x < 2 π

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