Asked by Temmick
                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.
            
        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.
Answers
                    Answered by
            Reiny
            
    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
<b>Ø = 0,π , 2π</b>
or
<b>Ø = π/3 , 5π/3</b>
    
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
<b>Ø = 0,π , 2π</b>
or
<b>Ø = π/3 , 5π/3</b>
                    Answered by
            Temmick
            
    Thanks soo much :)
    
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