What is the general solution for sin^4(x)-2sin^2(x)+1=0 ?

The general solution for sin^4(x)-2sin^2(x)+1=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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sin^4(x)-2sin^2(x)+1=0

sin^{4} (x) - 2 sin^{2} (x) + 1 = 0

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

Solution steps

sin^{4} (x) - 2 sin^{2} (x) + 1 = 0

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

- 2 sin (x) = 3

(\frac{2}{2}) (cos (θ) + sin (θ)) = 1

- 6 cos^{2} (θ) + 14 cos (θ) + 15 = 3 cos (θ) + 8

cos (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

2 sin (θ) = 3

What is the general solution for sin^4(x)-2sin^2(x)+1=0 ?
The general solution for sin^4(x)-2sin^2(x)+1=0 is x= pi/2+2pin,x=(3pi)/2+2pin