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

The general solution for -1+2sin^2(x)=0 is x= pi/4+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin,x=(7pi)/4+2pin

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

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

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solution steps

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

Solve by substitution

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

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

θ, y = - 2 + cos (3 θ)

cos (2 x) + 5 \cdot cos (x) = - 3

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

8 sin (x) + 7 cos (x) = 1

sec (y) = 12

What is the general solution for -1+2sin^2(x)=0 ?
The general solution for -1+2sin^2(x)=0 is x= pi/4+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin,x=(7pi)/4+2pin