What is the general solution for sin^2(2θ)=1,-180<= θ<= 180 ?

The general solution for sin^2(2θ)=1,-180<= θ<= 180 is θ=45,θ=-135,θ=135,θ=-45

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sin^2(2θ)=1,-180<= θ<= 180

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

Solution steps

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

Solve by substitution

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = - 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 13 5^{\circ}, θ = - 4 5^{\circ}

Combine all the solutions

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

cos (A) = \frac{1}{3}

100 = \frac{50}{sin ( \frac{x}{2} )}

tan (2 θ) = \frac{- 2}{1 - 1}

3 - 3 cos (x) = 3

cos (x) = \frac{5}{6}, \frac{3 π}{2} \leq x \leq 2 π

What is the general solution for sin^2(2θ)=1,-180<= θ<= 180 ?
The general solution for sin^2(2θ)=1,-180<= θ<= 180 is θ=45,θ=-135,θ=135,θ=-45