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

The general solution for sin(x)+2sin(x)cos(x)=0,0<= x<= 2pi is x=0,x=pi,x=2pi,x=(2pi)/3 ,x=(4pi)/3

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

sin (x) + 2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π, x = \frac{2 π}{3}, x = \frac{4 π}{3}

Solution steps

sin (x) + 2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

Factor

sin (x) + 2 sin (x) cos (x) : sin (x) (2 cos (x) + 1)

sin (x) (2 cos (x) + 1) = 0

Solving each part separately

sin (x) = 0 or 2 cos (x) + 1 = 0

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

2 cos (x) + 1 = 0, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{2 π}{3}, x = \frac{4 π}{3}

9 cos^{2} (θ) - 24 sin (θ) - 10 = - 8 sin (θ) + 6

cos^{2} (θ) = sin^{2} (θ) - 1

2 cos (2 θ) = - 3

10 cos^{2} (x) + 11 cos (x) + 3 = 0

4 csc (x) - 5 = 3

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