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    Hi!

    I'm re-doing some trigonometry basics in my spare time with a few A-level textbooks because I want to do some more studying in the future. Occasionally running into a few simple things I can't remember the explaination for. In this trig equation question:

    4sinx=3tanx, -180<x<180 which I rearrange to give cosx=3/4.

    I get the answers (+/-)41.41 degrees and 0 which are correct, but the textbook answer section also lists (+/-)180 degrees as an answer! I thought that the only way (+/-)180 would be an answer is if cosx had a negative value but I've run into this a few times and think I must have just forgotten why it happens.

    Thanks for any help you can provide!
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    (Original post by SGR2917)
    Hi!

    I'm re-doing some trigonometry basics in my spare time with a few A-level textbooks because I want to do some more studying in the future. Occasionally running into a few simple things I can't remember the explaination for. In this trig equation question:

    4sinx=3tanx, -180<x<180 which I rearrange to give cosx=3/4.

    I get the answers (+/-)41.41 degrees and 0 which are correct, but the textbook answer section also lists (+/-)180 degrees as an answer! I thought that the only way (+/-)180 would be an answer is if cosx had a negative value but I've run into this a few times and think I must have just forgotten why it happens.

    Thanks for any help you can provide!
    You lost this solution because you cancelled sin(x) off both sides, instead of factoring it out.
    latex in progress...
    4\sin(x)-3\tan(x)=0, \sin(x)(4-3\sec(x))=0, so \sin(x)=0 or 4-3\sec(x)=0, \cos(x)=\frac{3}{4}.
    done
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    (Original post by EricPiphany)
    You lost this solution because you cancelled sin(x) off both sides, instead of factoring it out.
    latex in progress...
    4\sin(x)-3\tan(x)=0, \sin(x)(4-3\sec(x))=0, so \sin(x)=0 or 4-3\sec(x)=0, \cos(x)=\frac{3}{4}.
    done
    That makes complete sense. Much appreciated! Thanks
 
 
 
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