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

The general solution for tan^2(x)+tan(x)=0,0<= x<2pi is x=0,x=pi,x=(3pi)/4 ,x=(7pi)/4

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

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

x = 0, x = π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

Solution steps

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

Solve by substitution

tan (x) = 0, tan (x) = - 1

tan (x) = 0, 0 \leq x < 2 π : x = 0, x = π

tan (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = 0, x = π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

csc (x) - 1 = 0

cos (2 x) + 5 cos (x) + 3 = 0

2 cos (θ) = 1

sin (x + \frac{π}{6}) - sin (x - \frac{π}{6}) = \frac{1}{2}

tan (θ) = \frac{5}{12}

What is the general solution for tan^2(x)+tan(x)=0,0<= x<2pi ?
The general solution for tan^2(x)+tan(x)=0,0<= x<2pi is x=0,x=pi,x=(3pi)/4 ,x=(7pi)/4