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 < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

Solve by substitution

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

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

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

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}

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

sin^{4} (x) - cos^{4} (x) = 0

tan (θ) cos (θ) + cos (θ) = 0

sin (x) = cos (x), 0 \leq x \leq 2 π

3 cos (θ) = 0

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