What is the general solution for 2sin^2(x)=1,+[0,+2pi] ?

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

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2sin^2(x)=1,+[0,+2pi]

2 sin^{2} (x) = 1, + [0, + 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, + 2 π]

Solve by substitution

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

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

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

3 sin (x) - 2 sin (x) cos (x) = 0

sin (x) = \frac{33}{80}

3 sin (x) + 5 = - 2 sin (x)

a, s (t) = tan (a t + π)

0 = sec (x) tan (x)

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