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

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

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

sin^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

sin^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

Solve by substitution

sin (x) = 1, sin (x) = - 1

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

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

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}

2 cos^{3} (θ) = cos (θ)

t, x = arccos (\frac{1}{1 + t ^{2}})

sin (x) \cdot cos (x) = sin (2 x)

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

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

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