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 \leq 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 \leq x < 2 π

Solve by substitution

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

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

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

arctan (x) = \frac{7.8}{- 4.42}

- sin (x + 5^{\circ}) = 0

cos (2 π x) \cdot 2 π + 2 π = 0

x, sin (y) = y cos (x)

4 sin (2 x) - 12 = 0, 0 \leq x \leq 2 π

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