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

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

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2sin^2(x)-2=0

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

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

Solution steps

2 sin^{2} (x) - 2 = 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 tan^{2} (x) = 13

2 tan^{2} (x) = 15

1 - cos (θ) = 1 + cos (θ)

cos (2 x) + cos (x) = 1

tan (x) sec (x) = - 2 tan (x)

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