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-/+ Root Solutions

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    \sqrt{10}cos(x-a)=2

    The mark scheme lists the only solutions when 2 is divided by positive root 10

    Why wouldn't the solutions where 2 is divided by negative root 10 be acceptable?
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    (Original post by pleasedtobeatyou)
    \sqrt{10}cos(x-a)=2

    The mark scheme lists the only solutions when 2 is divided by positive root 10

    Why wouldn't the solutions where 2 is divided by negative root 10 be acceptable?
     \sqrt4 = 2 \not= \pm 2

    The value obtained from the square root is always taken as positive.
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    maybe there is some limitation on x, like 0\leq x\leq \pi
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    This is from Core 4 but in some Core 3 questions for example,

    tan^2x = 6

    There's a need to find the solutions for -root6 and +root6
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    (Original post by njl94)
    maybe there is some limitation on x, like 0\leq x\leq \pi
    Nope, the solutions for the negative root would also be acceptable
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    (Original post by pleasedtobeatyou)
    This is from Core 4 but in some Core 3 questions for example,

    tan^2x = 6

    There's a need to find the solutions for -root6 and +root6
     \tan^2 x  = 6 \implies \tan x = \pm \sqrt{6}

    It is the same as  x^2 = 4 \implies x  = \pm \sqrt4 = \pm 2

    But  x = \sqrt4 = 2 \not= \pm \sqrt2

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