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    Hello guys,
    I'm struggling to find the value of x from a cos equation. i attached my working here please have a look and hint me if I'm wrong anywhere
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    (Original post by Alen.m)
    Hello guys,
    I'm struggling to find the value of x from a cos equation. i attached my working here please have a look and hint me if I'm wrong anywhere
    That looks correct. Also, \pi + \frac{\pi}{2}, 2\pi - \frac{\pi}{2} and \pi - \frac{\pi}{2} will be solutions to \cos f(x) = 0.
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    (Original post by Zacken)
    That looks correct. Also, \pi + \frac{\pi}{2}, 2\pi - \frac{\pi}{2} and \pi - \frac{\pi}{2} will be solutions to \cos f(x) = 0.
    well the reason I'm asking is because apparently the text book got cos^-1(0)=(pi) but mine is cos^-1(0)=(pi/2) which confused me bit and also my answer is 0.60 down of the text books answer
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    (Original post by Alen.m)
    well the reason I'm asking is because apparently the text book got cos^-1(0)=(pi) but mine is cos^-1(0)=(pi/2) which confused me bit and also my answer is 0.60 down of the text books answer
    cos^(-1) (0) = pi/2 is definitely correct, why don't you post the full question on here?
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    (Original post by Zacken)
    cos^(-1) (0) = pi/2 is definitely correct, why don't you post the full question on here?
    here's the full question which i've managed to fully done both parts a and b but for part c as i explained above because of that confusion i got an answer of 2.65 but the text book got it as 2.06
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    (Original post by Alen.m)
    here's the full question which i've managed to fully done both parts a and b but for part c as i explained above because of that confusion i got an answer of 2.65 but the text book got it as 2.06
    Of course! This is why you should be giving us the question.

    The minimum value of g(x) occurs when \cos (\text{whatever}) = -1, since -1 < 0. So you really want to solve \cos (x + 1.08) = -1
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    (Original post by Zacken)
    Of course! This is why you should be giving us the question.

    The minimum value of g(x) occurs when \cos (\text{whatever}) = -1, since -1 < 0. So you really want to solve \cos (x + 1.08) = -1
    I'm sorry but I'm still struggling , shouldn't we take cos^-1 of both sides of equation as i've attached for you before in order to find the value of x?
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    (Original post by Alen.m)
    I'm sorry but I'm still struggling , shouldn't we take cos^-1 of both sides of equation as i've attached for you before in order to find the value of x?
    The minimum value of your expression occurs when \cos (x + 1.08) = -1. So you need to take the inverse cosine of both sides of that equation, not yours.
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    (Original post by Zacken)
    The minimum value of your expression occurs when \cos (x + 1.08) = -1. So you need to take the inverse cosine of both sides of that equation, not yours.
    but if you see the text books answer part c which i attached you here the equation they used is equal to 0 not -1
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    (Original post by Alen.m)
    but if you see the text books answer part c which i attached you here the equation they used is equal to 0 not -1
    Apologies. I didn't notice the thing was square. Your answer is correct, the textbook is wrong.
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    (Original post by Zacken)
    That looks correct. Also, \pi + \frac{\pi}{2}, 2\pi - \frac{\pi}{2} and \pi - \frac{\pi}{2} will be solutions to \cos f(x) = 0.
    Lol you always come first when someone needs help with maths!
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    (Original post by Shiv is Light)
    Lol you always come first when someone needs help with maths!
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    (Original post by Zacken)
    It's ridiclous how they put wrong answers there lol Thanks anyway was struglling with it the whole day
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    (Original post by Shiv is Light)
    Lol you always come first when someone needs help with maths!
    This is how you beat him

    as soon as you see the question, you quote the OP with just 2 dots so you are first.
    then you think about the answer you are going to give and you edit the first post !
 
 
 
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