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

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

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

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

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

Solution steps

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

Solve by substitution

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

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

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

sin (3 θ) = 1, 0 \leq θ < 2 π

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

sin (2 x) cos (x) = sin (x)

tan (\frac{1}{2} x) - 1 = 0

sin (3 x) = 0, 0 \leq x \leq 2 π

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