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

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

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

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

x = \frac{π}{2}

Solution steps

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

Solve by substitution

sin (x) = 1, sin (x) = - 2

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

sin (x) = - 2, 0 \leq x < 2 π :

No Solution

Combine all the solutions

x = \frac{π}{2}

sin^{2} (x) + 5 sin (x) = 0

tan (α) = - \frac{4}{3}, \frac{π}{2} < α < π, cos (β) = \frac{1}{2}, 0 < β < \frac{π}{2}

- 3 cot (\frac{3 x}{2}) csc (\frac{3 x}{2}) = 0

sec (θ) = \frac{17}{8}

3 sin^{2} (x) = 1

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