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

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

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

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

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

Solution steps

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

Solve by substitution

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

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

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

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

5 sin (x) - 12 cos (x) = 13

sec (x) = 1.562641

cos (a) = - \frac{1}{2}

- 4 cos (2 x) + 4 sin (x) = 0

6 sin (θ) = 1

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