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    Solve the following equation for , in the interval :

     tan \theta = tan \theta (2+3sin \theta)

     \frac{tan \theta}{tan \theta} = 2 + 3sin \theta

     Sin \theta = -\frac {1}{3} With this I got 2 of the solutions.

    However, apparently 180^o & 360^o are solutions to?

    What did I do wrong?
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    (Original post by Naruke)
    Solve the following equation for , in the interval :

     tan \theta = tan \theta (2+3sin \theta)

     \frac{tan \theta}{tan \theta} = 2 + 3sin \theta

     Sin \theta = -\frac {1}{3} With this I got 2 of the solutions.

    However, apparently 180^o & 360^o are solutions to?

    What did I do wrong?
    You cannot divide by  tan \theta because you do not know what it is equal to - it could be equal to zero.

    Try expanding the brackets on the right and then bringing everything on one side of the equation.
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    (Original post by MathQ123)
    You cannot divide by  tan \theta because you do not know what it is equal to - it could be equal to zero.

    Try expanding the brackets on the right and then bringing everything on one side of the equation.
    Thanksssssssssssssssss
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    (Original post by Naruke)
    Thanksssssssssssssssss
    It's kind of analogous to when you have x = x(x+2), you'd normally move everything to one side to get: x(x+2) - x = 0 \Rightarrow x(x+1) = 0 which gives you two solutions. You recognise that it is folly to do x = x(x+2) \Rightarrow \frac{x}{x} = x+2 \Rightarrow x+2 = 1, right?

    Same thing here, never cancel trigonometric functions, always factorise them out. So in this case:

    \tan \theta = \tan \theta(2 + 3\sin\theta) \Rightarrow \tan \theta(2 + 3\sin \theta) - \tan \theta = 0 \Rightarrow \tan \theta(1 + 3\sin\theta) = 0
 
 
 
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