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

The general solution for 2sin^2(x)+sin(x)=0,0<= x<= 2pi is x=0,x=pi,x=2pi,x=(7pi)/6 ,x=(11pi)/6

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

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

x = 0, x = π, x = 2 π, x = \frac{7 π}{6}, x = \frac{11 π}{6}

Solution steps

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

Solve by substitution

sin (x) = 0, sin (x) = - \frac{1}{2}

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{7 π}{6}, x = \frac{11 π}{6}

6 cos^{2} (x) - 5 cos (x) = 4

cos (2 0^{\circ}) = sin (x)

2 cos (x) + sec (x) = 0

sec (2 x) = 1

2 cos^{2} (x) + sin^{2} (x) = 2

What is the general solution for 2sin^2(x)+sin(x)=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)+sin(x)=0,0<= x<= 2pi is x=0,x=pi,x=2pi,x=(7pi)/6 ,x=(11pi)/6