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    Find in radians correct to 2 decimal places, the two smallest positive values of θ for which sin(2θ +1/3π) = 0.123. I've worked out 2.68 but I figure out how to get to the other answer.
    If you could help that would be great.
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    Hi, these are the steps I used:
    (I'll have to use 'a' for theta as I cannot get that symbol)

    sin(2a+1/3pi)=0.123
    2a+1/3pi=sin^-1(0.123)
    2a+1/3pi=0.123
    2a=-0.925
    a=-0.462

    You then use the symmetry of the sin(a) graph:
    (-pi)+0.462= -2.68

    I hope two things: that's correct and that it helps you if it is. I cannot see how you got +2.68 (unless there is a typo in the original message or my working out)
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    (Original post by Jasminea)
    Find in radians correct to 2 decimal places, the two smallest positive values of θ for which sin(2θ +1/3π) = 0.123. I've worked out 2.68 but I figure out how to get to the other answer.
    If you could help that would be great.
    If you call x = 2\theta + \frac{\pi}{3} and then solve \sin x = 0.123 then we have x=\sin^{-1} 0.123 + 2\pi and then x = \pi - \sin^{-1} (0.123) +2\pi.

    Back-sub: 2\theta+ \frac{\pi}{3} = \sin^{-1} 0.123 + 2\pi (solve for theta) for the first solution and 2\theta + \frac{\pi}{3} =  \pi - \sin^{-1} (0.123) +2\pi for the second solution.
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    (Original post by ErraticPhysicist)
    Hi, these are the steps I used:
    (I'll have to use 'a' for theta as I cannot get that symbol)

    sin(2a+1/3pi)=0.123
    2a+1/3pi=sin^-1(0.123)
    2a+1/3pi=0.123
    2a=-0.925
    a=-0.462

    You then use the symmetry of the sin(a) graph:
    (-pi)+0.462= -2.68

    I hope two things: that's correct and that it helps you if it is. I cannot see how you got +2.68 (unless there is a typo in the original message or my working out)
    It asks for positive values.
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    (Original post by Zacken)
    It asks for positive values.
    Ah yes, I see. Your answer looks correct, well done
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    (Original post by Zacken)
    If you call x = 2\theta + \frac{\pi}{3} and then solve \sin x = 0.123 then we have x=\sin^{-1} 0.123 + 2\pi and then x = \pi - \sin^{-1} (0.123) +2\pi.

    Back-sub: 2\theta+ \frac{\pi}{3} = \sin^{-1} 0.123 + 2\pi (solve for theta) for the first solution and 2\theta + \frac{\pi}{3} =  \pi - \sin^{-1} (0.123) +2\pi for the second solution.
    Ok I can solve it but how did you get 2Pi? Other than that, thanks for help.
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    (Original post by Jasminea)
    Ok I can solve it but how did you get 2Pi? Other than that, thanks for help.
    Well, trigonometric functions are two-pi periodic, i.e: they repeat themselves after every two pi radians.

    So if x is a solution to a trig equation, then so is x + 2pi, so is x + 2pi + 2pi, so is x + 2pi + 2pi + 2pi, etc... because sin (x) = sin (x + 2pi) since the function just repeats itself after 2pi, so if I add 2pi, it changes nothing.
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    Ok. Yeah that makes sense thanks.
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    (Original post by Jasminea)
    Ok. Yeah that makes sense thanks.
    No problem.
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    (Original post by Zacken)
    No problem.
    Hi, did u ever get 0.99 as a value, bcoz that's the other answer in the back of my textbook? I can't get that answer. It may be wrong.
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    (Original post by Jasminea)
    Hi, did u ever get 0.99 as a value, bcoz that's the other answer in the back of my textbook? I can't get that answer. It may be wrong.
    It's nor a mistake.

    \displaystyle 2\theta + \frac{\pi}{3} = \frac{\pi}{2} - \sin^{-1} (0.123).
 
 
 
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