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    (sinx -1) (5cosx + 3) = 0

    Solve the equation for x, in the interval 0 < x < 360

    What I have done:

    (sinx -1) (5cosx + 3) = 0

    √cos^2 x + √sin^2 x = √1 - Is this step correct? I just square rooted everything in the identity - I think it's correct, but just want to make sure.

    ((1 - cosx) -1) (5cosx + 3) = 0
    -cosx(5cosx + 3) = 0
    -5cos^2x - 3cosx = 0
    cosx(-5cosx - 3) = 0
    cos x = 0 or cos x = -3/5

    x = 90, x = 126.9 (the first values are obtained by finding the inverse cos of 0 and -3/5), x = 233.1, x = 270 (the second set of values are obtained by finding 360-x, since cos x = cos (360-x))

    I'm not sure where I may have gone wrong. I did get the first 3 values of x right: x = 90, x = 126.9, x = 233.1. However, x = 270 is incorrect.

    I have graphed y = (sinx -1) (5cosx + 3), and there are 3 solutions at x = 90, x = 126.9, x = 233.1.

    I also graphed y = cosx(-5cosx - 3), and there are 4 solutions at x = 90, x = 126.9, x = 233.1, x = 270.

    It's just that last x = 270 value that's annoying me - I can't seem to work out what's gone wrong and where it came from.

    Thank you very much!
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    (Original post by zebra015)
    (sinx -1) (5cosx + 3) = 0

    Solve the equation for x, in the interval 0 < x < 360

    What I have done:

    (sinx -1) (5cosx + 3) = 0

    √cos^2 x + √sin^2 x = √1 - Is this step correct? I just square rooted everything in the identity - I think it's correct, but just want to make sure.

    ((1 - cosx) -1) (5cosx + 3) = 0
    -cosx(5cosx + 3) = 0
    -5cos^2x - 3cosx = 0
    cosx(-5cosx - 3) = 0
    cos x = 0 or cos x = -3/5

    x = 90, x = 126.9 (the first values are obtained by finding the inverse cos of 0 and -3/5), x = 233.1, x = 270 (the second set of values are obtained by finding 360-x, since cos x = cos (360-x))

    I'm not sure where I may have gone wrong. I did get the first 3 values of x right: x = 90, x = 126.9, x = 233.1. However, x = 270 is incorrect.

    I have graphed y = (sinx -1) (5cosx + 3), and there are 3 solutions at x = 90, x = 126.9, x = 233.1.

    I also graphed y = cosx(-5cosx - 3), and there are 4 solutions at x = 90, x = 126.9, x = 233.1, x = 270.

    It's just that last x = 270 value that's annoying me - I can't seem to work out what's gone wrong and where it came from.

    Thank you very much!


    You're making it waaaay more difficult for yourself.

    (sinx -1) ---> sinx = 1, hence x = 90.
    (5cosx + 3) ---> cosx = -3/5, hence x = 126.9 and (360 - 126.9 =) 233.1.

    That's it. No need to substitute any identities.
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    Without going further, I can tell you that you can't square root the identity the way you did. The square root sign doesn't separate out, so instead it would look like this:

    \sqrt{cos^{2}x + sin^{2}x} = \sqrt{1}

    Furthermore, your equation is already factorised so you can form two equations,

    sinx = 1

cosx = -3/5
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    (Original post by zebra015)
    (sinx -1) (5cosx + 3) = 0

    Solve the equation for x, in the interval 0 < x < 360

    What I have done:

    (sinx -1) (5cosx + 3) = 0

    √cos^2 x + √sin^2 x = √1 - Is this step correct? I just square rooted everything in the identity - I think it's correct, but just want to make sure.

    ((1 - cosx) -1) (5cosx + 3) = 0
    -cosx(5cosx + 3) = 0
    -5cos^2x - 3cosx = 0
    cosx(-5cosx - 3) = 0
    cos x = 0 or cos x = -3/5

    x = 90, x = 126.9 (the first values are obtained by finding the inverse cos of 0 and -3/5), x = 233.1, x = 270 (the second set of values are obtained by finding 360-x, since cos x = cos (360-x))

    I'm not sure where I may have gone wrong. I did get the first 3 values of x right: x = 90, x = 126.9, x = 233.1. However, x = 270 is incorrect.

    I have graphed y = (sinx -1) (5cosx + 3), and there are 3 solutions at x = 90, x = 126.9, x = 233.1.

    I also graphed y = cosx(-5cosx - 3), and there are 4 solutions at x = 90, x = 126.9, x = 233.1, x = 270.

    It's just that last x = 270 value that's annoying me - I can't seem to work out what's gone wrong and where it came from.

    Thank you very much!
    Firstly. Think of this as factorising a equation. You will get solutions by equating the factors to 0.

    So to find the solutions:

    Equate them to 0...
    Sin x = 1 and cos x = -3/5
    Solve using that as simple as that !

    Also, your square root is wrong. When you square root an equation you square root both sides. You've square rooted each term individually which is wrong. √(Cos^2 x + sin^2 x) is not the same as
    √cos^2 x + √sin^2 x....

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    (Original post by zebra015)
    ...
    Others have said it but I want to jump on this train. So, why not just say \sin(x)-1=0 or 5\cos(x)+3=0 ??
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    √( 9 + 16 ) should = 5

    if you try

    √9 + √16 you get 7....
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    (Original post by RDKGames)
    Others have said it but I want to jump on this train. So, why not just say \sin(x)-1=0 or 5\cos(x)+3=0 ??

    FML i didnt even realise it was factorised :unimpressed::unimpressed::unimpressed::laugh::laugh::laugh: This would've made it a lot easier ..... + rep to all

    thank you all for ur help!!
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    (Original post by the bear)
    √( 9 + 16 ) should = 5

    if you try

    √9 + √16 you get 7....
    Always the best . Got to rep you xD


    Posted from TSR Mobile
 
 
 
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