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    solve the equation sec θ = 5 cosec θ for 0 ≤ θ ≤ 2π

    I don't know if I'm just confusing myself or this question is hard please help if you can
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    (Original post by nazz97)
    solve the equation sec θ = 5 cosec θ for 0 ≤ θ ≤ 2π

    I don't know if I'm just confusing myself or this question is hard please help if you can
    How would you go about solving this?

    (for a lot of trig questions It may help to rewrite everything in terms of sin and cos)
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    (Original post by nazz97)
    solve the equation sec θ = 5 cosec θ for 0 ≤ θ ≤ 2π

    I don't know if I'm just confusing myself or this question is hard please help if you can
    Convert them into sin and cos theta, simplify and divide both sides by cos theta. You should get an equation in tan.
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    (Original post by ubisoft)
    Convert them into sin and cos theta, simplify and divide both sides by cos theta. You should get an equation in tan.
    Or just multiply both sides by sin(theta) and solve...
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    (Original post by ubisoft)
    Convert them into sin and cos theta, simplify and divide both sides by cos theta. You should get an equation in tan.
    If you're dividing by \cos \theta should you not account for the fact that \cos \theta = 0?
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    (Original post by Zacken)
    If you're dividing by \cos \theta should you not account for the fact that \cos \theta = 0?
    You're right, that could occur. You should factorise cosx instead.
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    (Original post by Zacken)
    If you're dividing by \cos \theta should you not account for the fact that \cos \theta = 0?
    Isn't it obvious that you should just multiply by sin(theta) and that's it?
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    (Original post by ubisoft)
    You're right, that could occur. You should factorise cosx instead.
    Dude, just multiply both sides by sin(theta), simple.
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    (Original post by PrimeLime)
    Isn't it obvious that you should just multiply by sin(theta) and that's it?
    Possibly, I haven't looked at the question. Just saw the post about dividing by cos and replied to that. Looking at it, I would multiply by sin theta as well and factorise; but it's okay if other people choose a different way to approach it. :-) (as long as it's valid and doesn't lose you solutions)
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    (Original post by Zacken)
    Possibly, I haven't looked at the question. Just saw the post about dividing by cos and replied to that. Looking at it, I would multiply by sin theta as well and factorise; but it's okay if other people choose a different way to approach it. :-) (as long as it's valid and doesn't lose you solutions)
    And factorise? If you multiply by sin(theta) you just get tan(theta)=5.
    Don't think it's necessary to multiply everything through and factorise cos(theta) etc.
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    (Original post by Zacken)
    If you're dividing by \cos \theta should you not account for the fact that \cos \theta = 0?

    (Original post by ubisoft)
    You're right, that could occur. You should factorise cosx instead.
    In this circumstance you can get away with dividing by cosx, since cosx=0 can't be a solution, as sec x would have to be infinite.
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    (Original post by PrimeLime)
    And factorise? If you multiply by sin(theta) you just get tan(theta)=5.
    Don't think it's necessary to multiply everything through and factorise cos(theta) etc.
    Ah, indeed, I didn't read the question properly, I thought it was asec x = bcosec x + c
    Yeah, you'd just get tan, you're right!
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    (Original post by ghostwalker)
    In this circumstance you can get away with dividing by cosx, since cosx=0 can't be a solution, as sec x would have to be infinite.
    Ah, that's true, thanks for that insight. That'll teach me to put pen to paper instead of talking crap.
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    (Original post by Zacken)
    Ah, indeed, I didn't read the question properly, I thought it was asec x = bcosec x + c
    Yeah, you'd just get tan, you're right!
    Yeah I've noticed a lot of people that dive into converting to cosines/sines and multiplying everything through mechanically without thinking a bit to see if there are quicker ways (not talking about you ). Sadly, this kind of thought is very common with most A-Level maths students.
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    (Original post by PrimeLime)
    Yeah I've noticed a lot of people that dive into converting to cosines/sines and multiplying everything through mechanically without thinking a bit to see if there are quicker ways (not talking about you ). Sadly, this kind of thought is very common with most A-Level maths students.
    There was a thread that came up with something like this (the way students learn methods instead of understanding) a few days ago, except it was in the context of integration. Can't remember it for the life of me though.
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    (Original post by Zacken)
    There was a thread that came up with something like this (the way students learn methods instead of understanding) a few days ago, except it was in the context of integration. Can't remember it for the life of me though.
    Speaking of integration, it's pretty shameful that schools don't even teach what integration even IS. They just say it's the reverse of differentiation. Oh, and it's also the area under a curve. And then they get you to do loads of mindless calculation.
    I think students don't learn nearly enough understanding - I had to get all of mine from extra reading. It's pretty bad how most of A-Level is just an exercise in algebra...
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    thanks guys i can't believe that i was stuck in this question!! just to check :
    1/ i wrote everything down in the form sin and cos
    2/ then i multiplied everything by sin theta
    3/ as a result i ended up with tan (theta)=5
    4/ found the values of tan theta in degrees and got (1.373 and 4.514) correct to 3 decimal places
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    (Original post by nazz97)
    thanks guys i can't believe that i was stuck in this question!! just to check :
    1/ i wrote everything down in the form sin and cos
    2/ then i multiplied everything by sin theta
    3/ as a result i ended up with tan (theta)=5
    4/ found the values of tan theta in degrees and got (1.373 and 4.514) correct to 3 decimal places
    Those are radians.
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    (Original post by nazz97)
    thanks guys i can't believe that i was stuck in this question!! just to check :
    1/ i wrote everything down in the form sin and cos
    2/ then i multiplied everything by sin theta
    3/ as a result i ended up with tan (theta)=5
    4/ found the values of tan theta in degrees and got (1.373 and 4.514) correct to 3 decimal places
    Number 4... why have you got it in radians? Does it ask for it in degrees or radians?
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    (Original post by Marxist)
    Number 4... why have you got it in radians? Does it ask for it in degrees or radians?
    The bounds given in the question are in radians, I think it'd be safe to assume radians will be used.
 
 
 
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