Solving Sin(2x)+Sin(x)=0 over [0,2π)

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To solve the equation sin(2x) + sin(x) = 0 over the interval [0, 2π), the equation can be rewritten as 2sin(x)cos(x) + sin(x) = 0. This leads to two cases: 2cos(x) = -1 and sin(x) = 0. The solutions for cos(x) = -0.5 are x = 2π/3 and 4π/3, while sin(x) = 0 gives x = 0 and π. Therefore, the complete set of solutions is x = 0, 2π/3, π, and 4π/3.
RansidMeat
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hello there,
how do i solve over [0,2pi) for sin(2x)+sin(x)=0
thanks for what help you can give
RansidMeat
 
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Yay, problem solved.

cookiemonster
 
RansidMeat said:
hello there,
how do i solve over [0,2pi) for sin(2x)+sin(x)=0
thanks for what help you can give
RansidMeat
sin(2x) + sin(x) = 0
2sin(x)cos(x) + sin(x) = 0
2sin(x)cos(x) = -sin(x)
2cos(x) = -1 OR sin(x) = 0
cos(x) = -0.5 OR x = 0, pi
x = 2pi/3, 4pi/3 OR x = 0, pi
x = 0, (2/3)pi, pi, (4/3)pi.
 
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