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    Ive been trying to finish c3 trig and im still having problems with the question i often don't no where to start and when i do i end up making the problem more complex then it was to begin.

    May i have some tips for asnwering c3 trig and what i should be looking for in the questions especial ones to do with cosesc, cot and sec.
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    When I'm solving trig equations I always do this:

    1.) Make every term the same. What I mean by this is, if you have cos2x, sinx or tan3x you want to make them all the same so in each trig function you just have x, 2x or 3x.

    2.) Make it terms of only one trig function. So by using identities, make it into one of the sine, cosine or tan functions (Preferably not sec, cosec or cot because they're harder to solve.)

    3.) Remember not to cancel trig functions in doing steps 1.) and 2.)! E.g. if you have  sinxcos3x = cosec2xsinx or sinx(1 - tanx) = 0 it is very tempting to cancel the sinx but DO NOT! They hold extra solutions.

    4.) Once you have something in the form  cos(ax) = b or something similiar you can solve it by the ACTS quadrant method or by looking at the graph of the trig function.
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    (Original post by soutioirsim)
    When I'm solving trig equations I always do this:

    1.) Make every term the same. What I mean by this is, if you have cos2x, sinx or tan3x you want to make them all the same so in each trig function you just have x, 2x or 3x.

    2.) Make it terms of only one trig function. So by using identities, make it into one of the sine, cosine or tan functions (Preferably not sec, cosec or cot because they're harder to solve.)

    3.) Remember not to cancel trig functions in doing steps 1.) and 2.)! E.g. if you have  sinxcos3x = cosec2xsinx or sinx(1 - tanx) = 0 it is very tempting to cancel the sinx but DO NOT! They hold extra solutions.

    4.) Once you have something in the form  cos(ax) = b or something similiar you can solve it by the ACTS quadrant method or by looking at the graph of the trig function.
    Im going to do this thanks, il keep this thread up to date with my progress if i get really stck again, thank you.
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    What about this one,  cosec\frac{1}{2}\theta=\frac{2 \sqrt{3}}{3}
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    I was just doing it. thanks though this is like when i was on log i couldn't do it for ages and wasn't able to see what everyone esle was seeing. also shouldnt that be  sin\frac{1}{2}\theta=\frac{3}{2 \sqrt{3}}
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    (Original post by Core)
    I was just doing it. thanks though this is like when i was on log i couldn't do it for ages and wasn't able to see what everyone esle was seeing. also shouldnt that be  sin\frac{1}{2}\theta=\frac{3}{2 \sqrt{3}}
    They're the same.

    Also, to solve it, you'll need some sort of range in which \theta lies in. Other than that, you just solve it as usual, with your acute angle, CAST diagrams, or whatever way you usually work them out.
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    (Original post by dknt)
    They're the same.

    Also, to solve it, you'll need some sort of range in which \theta lies in. Other than that, you just solve it as usual, with your acute angle, CAST diagrams, or whatever way you usually work them out.
    K thanks the book gave the range and i worked out the correct answer, thanks to all for the all the help on this thread.
 
 
 
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