What is the general solution for sin^4(x)=sin^2(x),0<= x<2pi ?

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

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sin^4(x)=sin^2(x),0<= x<2pi

sin^{4} (x) = sin^{2} (x), 0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

Solution steps

sin^{4} (x) = sin^{2} (x), 0 \leq x < 2 π

Solve by substitution

sin (x) = 1, sin (x) = - 1, sin (x) = 0

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}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

sin (θ) - 0.1 cos (θ) = \frac{6.88}{9.8}

sin (x) = 0.3352

tan^{2} (3 y) - 2 sec (3 y) - 2 = 0

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

sin^{4} (x) = - \frac{1}{8}

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