What is the general solution for sin(2x)+sin(x)=0,0<= x<= 360 ?

The general solution for sin(2x)+sin(x)=0,0<= x<= 360 is x=0,x=180,x=360,x=120,x=240

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sin(2x)+sin(x)=0,0<= x<= 360

sin (2 x) + sin (x) = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 0, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Solution steps

sin (2 x) + sin (x) = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Rewrite using trig identities

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

Factor

sin (x) + 2 cos (x) sin (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 36 0^{\circ} : x = 0, x = 18 0^{\circ}, x = 36 0^{\circ}

2 cos (x) + 1 = 0, 0 \leq x \leq 36 0^{\circ} : x = 12 0^{\circ}, x = 24 0^{\circ}

Combine all the solutions

x = 0, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

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

cot (x) = - 3

csc (\frac{2 π}{3})

cos (x) = \frac{1}{4}

cot (\frac{3 π}{4})

What is the general solution for sin(2x)+sin(x)=0,0<= x<= 360 ?
The general solution for sin(2x)+sin(x)=0,0<= x<= 360 is x=0,x=180,x=360,x=120,x=240