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

The general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<2pi is x= pi/2 ,x= pi/6 ,x=(5pi)/6

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2sin^2(x)-3sin(x)+1=0,0<= x<2pi

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

x = \frac{π}{2}, x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

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

Solve by substitution

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

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

sin (x) = \frac{1}{2}, 0 \leq x < 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

x = \frac{π}{2}, x = \frac{π}{6}, x = \frac{5 π}{6}

cos (x) = \frac{3}{10}

5 cos (3 x + 60) = 4

3 sin^{2} (x) = 0

\frac{5 3}{3} cos (x) = 1 + 3 cos (x)

sin (x) n = 3

What is the general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<2pi ?
The general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<2pi is x= pi/2 ,x= pi/6 ,x=(5pi)/6