What is the general solution for 2sin^2(θ)-1=0,0<= θ<2pi ?

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

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2sin^2(θ)-1=0,0<= θ<2pi

2 sin^{2} (θ) - 1 = 0, 0 \leq θ < 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}, θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Solution steps

2 sin^{2} (θ) - 1 = 0, 0 \leq θ < 2 π

Solve by substitution

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

sin (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{4}, θ = \frac{3 π}{4}

sin (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Combine all the solutions

θ = \frac{π}{4}, θ = \frac{3 π}{4}, θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

2 cosh (2 x) = 5 sinh (x) + 3

tan (x) = \frac{6.9}{6.6}

sin (x) - cos (x) = \frac{2}{2}

2 cos (x) = 7 - \frac{3}{cos ( x )}

- sin (x) + 2 = 0

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