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    so I'm having problems answering this question:

    Solve, for –180° ≤ X < 180°

    (1 + tan X)(5 sin X – 2) = 0
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    Tanx = sinx/cosx
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    Then multiply out
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    What happens if  \tan x =-1 and/or  \sin x =\frac{2}{5} .

    So we have  \tan x = -1 \Rightarrow x=\arctan(-1) <--- standard angle.

    Also  \sin x= \frac{2}{5} \Rightarrow x= \arcsin \left ( \frac{2}{5} \right ) .
    Use the fact that tan has a period of  \pi and sin has a period of  2\pi to find all the answers in the specified range.
    The question is almost solved for you, there is no manipulation required all you need to do is what I did above.
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    Might also have to use sin^2 + cos^2 = 1 and rearranged
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    Do the inverse as there are solutions for tan, cos and sin between -1 and 1
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    (Original post by audreygal)
    Then multiply out
    this is not necessary...

    if A x B = 0

    then either A = 0 or B = 0
 
 
 
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