Finding Solutions for Sin 2θ = Sin θ in the Interval 0 ≤ θ ≤ 360?

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SUMMARY

The equation sin 2θ = sin θ can be solved by utilizing the identity sin 2θ = 2sinθcosθ. By rearranging the equation to 2sinθcosθ - sinθ = 0, it can be factored to sinθ(2cosθ - 1) = 0. This results in two sets of solutions: sinθ = 0, yielding θ = 0, 180, 360 degrees, and 2cosθ - 1 = 0, yielding θ = 60 and 300 degrees. Therefore, the complete set of solutions in the interval 0 ≤ θ ≤ 360 is θ = 0, 60, 180, 300, and 360 degrees.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sin 2θ = 2sinθcosθ
  • Knowledge of solving trigonometric equations
  • Familiarity with the unit circle and angle measures in degrees
  • Ability to factor algebraic expressions
NEXT STEPS
  • Study the derivation and applications of trigonometric identities
  • Practice solving various trigonometric equations
  • Explore the unit circle and its significance in trigonometry
  • Learn about the graphical representation of sine functions
USEFUL FOR

Students studying trigonometry, educators teaching trigonometric equations, and anyone looking to enhance their understanding of solving sine equations within specified intervals.

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Homework Statement



Find all values of θ in the interval 0 ≤ θ ≤ 360 that satisfy the equation sin 2θ = sin θ

Homework Equations



sin2θ = 2sinAcosA

The Attempt at a Solution



I'm a little confused on how to start off knowing sin2θ is equivalent to 2sinAcosA, would you factor this?
 
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sin2A = SinA
2SinAcosA = SinA
2SinAcosA - SinA = 0
SinA (2CosA-1)= 0
sinA = 0 2CosA-1=0
A= 0,180,360 A= 60, 300

i didn't check this, i just did this real quick, if i was wrong sorry, but i think i did it right
Hope that helps
 


Thanks a lot for this :)
 

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