Solving Sin(2x)+Sin(x)=0 over [0,2π)

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SUMMARY

The equation sin(2x) + sin(x) = 0 can be solved over the interval [0, 2π). The solution involves transforming the equation into 2sin(x)cos(x) + sin(x) = 0, which simplifies to 2cos(x) = -1 or sin(x) = 0. The definitive solutions are x = 0, π, 2π/3, and 4π/3.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sin(2x) and sin(x).
  • Knowledge of solving trigonometric equations.
  • Familiarity with the unit circle and angle measures in radians.
  • Basic algebraic manipulation skills.
NEXT STEPS
  • Study the derivation of trigonometric identities, focusing on double angle formulas.
  • Learn how to graph sine functions to visualize solutions over specific intervals.
  • Explore advanced techniques for solving trigonometric equations, including factoring and substitution methods.
  • Investigate periodic properties of trigonometric functions to understand solution sets beyond the primary interval.
USEFUL FOR

Students studying trigonometry, mathematics educators, and anyone interested in solving trigonometric equations effectively.

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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