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    Can someone tell me how to approach this question (its in AEA 04 )

    1) cosx + √ (1-0.5sin2x) = 0

    Thanks!
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    what's the domain?
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    (Original post by darkenergy)
    Can someone tell me how to approach this question (its in AEA 04 )

    1) cosx + √ (1-0.5sin2x) = 0

    Thanks!
    You could square both sides but crucially you need to remember you lose the signs.
    Alternatively you could multiply both sides by \cos x - \sqrt{1-0.5\sin 2x} to give:
    ( \cos x + \sqrt{1-0.5\sin 2x} )( \cos x - \sqrt{1-0.5\sin 2x} )=0 \\

cos^{2}x-(1-0.5\sin 2x) = 0
    but you need to remember \cos x \neq \sqrt{1-0.5\sin 2x}
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    I just plotted the graph, and only get 1 solution...2
    How is this possible, if there are 3 terms in the eq?
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    (Original post by Goldenratio)
    I just plotted the graph, and only get 1 solution...2
    How is this possible, if there are 3 terms in the eq?
    It's defined over (0,360). It's not a polynomial so I don't think there are any explicit rules about how many roots it might have that we should know of.
    I get 2 roots. Try solve it algebraically.
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    what if the equation becomes

    cosx + √(1-0.5sin2x) + 2 = 0

    is this solvable?
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    (Original post by darkenergy)
    what if the equation becomes

    cosx + √(1-0.5sin2x) + 2 = 0

    is this solvable?
    I don't think that has solutions.
    You're saying: cosx+√(1-0.5sin2x)=-2
    but √(1-0.5sin2x)>=0 so you need cosx<=-2.
    But |cosx|<=1 so no solutions exist.
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    what if it is:

    cosx+√(1-0.5sin2x)= 2?

    I am just wondering how to solve such equations when there is a number term so you can just square both sides
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    I started with your lead but i can't get rid of the -1.
    First so i did:
    cos^2x-(1-1/2sin2x)=0
    => cos^2x-1+sinxcosx=0
    therefore
    cosx(cosx+sinx)=1
    cosx+sinx=secx
    square both sides so: cosx^2+2sinxcosx+sinx^2=sec^2x
    1+sin2x=sec^2x
    2sinxcosx=tan^2x
    sin2x=tan^2x but then realised it just complicated it...
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    (Original post by darkenergy)
    what if it is:

    cosx+√(1-0.5sin2x)= 2?

    I am just wondering how to solve such equations when there is a number term so you can just square both sides
    Let's try:
    √(1-0.5sin2x)=(2-cosx).
    Evidently both sides are greater than 0 for all x so you can square without adding spurious solutions.
    1-0.5sin2x=4-4cosx+(cosx)^2
    (cosx)^2-4cosx+3+0.5sin2x=0
    (cosx)^2-4cosx+3+sinxcosx=0
    c^2-4c+3+c√(1-c^2)=0
    It clearly has solutions (eg x=0) but it's not easy to solve exactly.
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    I'd do it in the following way:
    \cos ^2 x -1+\frac{1}{2} sin2x = 0  \\

\cos 2x + 1 - 2 + \sin 2x=0 \\

\sin 2x + \cos 2x = 1 \\

R\cos (2x+\alpha)= 1
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    Ah! yep nice one!
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    k thanks people
 
 
 
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