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

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

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

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

x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

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

Solve by substitution

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

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

sin (x) = - 1, 0 < x < π :

No Solution

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = - \frac{4}{5}

5 sin (2 x) sin (x) = 5 cos (x)

(tan (x) - 1) (cos (x) + 1) = 0

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

cos (2 x - \frac{π}{6}) = \frac{3}{2}

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