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

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

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2sin^2(x)-1=0,0<= x<2pi

2 sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

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

Solution steps

2 sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

Solve by substitution

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

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

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

Combine all the solutions

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

tan^{2} (x) + 2 tan (x) = - 1

4 csc^{2} (x) - 2 = 0

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

18 cos (x) - 93 = 0

2 sin (3 θ) - 1 = 0

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