Find the values of theta (0,2pi) that satisfy sin theta= square root 3 over 2

Solve 2costheta sintheta-sintheta on (0, 2pi)

please I really need help in knowing how to solve these questions. quite urgent. I don't understand how to do them at all.

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  1. You should recognize √3/2 to be a ratio from the 30-60-90° right angled triangle.
    e.g. I know sin 60° = √3/2
    and 60° = π/3 radians
    I also know that the sine is positive in quadrants I and II by the CAST rule,
    so Ø = π/3 and π - π/3 or 2π/3 , .......( 60° and 120°)

    Ø = π/3 , 2π/3

    I assume your second is
    2cosØsinØ - sinØ = 0
    it factors ...
    sinØ(2cosØ - 1) = 0
    sinØ = 0 or cosØ = 1/2
    Ø = 0,π , 2π
    Ø = π/3 , 5π/3

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    posted by Reiny
  2. Thanks soo much :)

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    posted by Temmick

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