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    4 cos(θ + 60◦) cos(θ + 30◦) ≡ √3 − 2 sin 2θ

    Given that there are no values of θ which satisfy the equation4 cos(θ + 60◦) cos(θ + 30◦) = k,determine the set of values of the constant k.

    --
    I've made sin2θ -1 and 1,
    So the minimum of the graph at sin2θ=-1 gives √3+2 and the maximum of the graph at sin2θ= 1 gives √3-2

    So for there to be no values k<√3+2 and k>√3-2
    But this is wrong on the mark scheme - they're meant to be the other way round.

    Where have I gone wrong?
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    draw a sketch. it's pretty obvious why your answer can't possibly be right
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    (Original post by CancerousProblem)
    draw a sketch. it's pretty obvious why your answer can't possibly be right
    I have, but i'm still confused.
    When sin2θ=-1, k must be root3 +2. The question says there's no solution so shouldn't it be k<root3 +2 at the maximum? (Where there is no sine wave)
    And when sin
    2θ=1, k>root3-2 at the minimum?



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    Look
    There are no values of θ that satisfy 4 cos(θ + 60◦) cos(θ + 30◦) = k
    is equivalent to no values of
    θ that satisfy √3 − 2 sin 2θ = k
    Therefore you are looking for the values of k OUTSIDE the range of the function
    Think of the range of values
    √3 − 2 sin 2θ can take. k is everything it CAN'T TAKE.
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    (Original post by Peanut247)
    I have, but i'm still confused.
    When sin2θ=-1, k must be root3 +2. The question says there's no solution so shouldn't it be k<root3 +2 at the maximum? (Where there is no sine wave)
    And when sin
    2θ=1, k>root3-2 at the minimum?
    You're using "minimum" and "maximum" a bit sloppily, which might be the problem. The maximum of \sin(2\theta) is 1, yes, but that makes the *minimum* of \sqrt{3}-2\sin(2\theta). (Because of the extra negative before the sin term.)
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    (Original post by Smaug123)
    You're using "minimum" and "maximum" a bit sloppily, which might be the problem. The maximum of \sin(2\theta) is 1, yes, but that makes the *minimum* of \sqrt{3}-2\sin(2\theta). (Because of the extra negative before the sin term.)
    Don't you factor in that negative when you sub -1 and +1 for sinθ into the equation?
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    (Original post by CancerousProblem)
    Look
    There are no values of θ that satisfy 4 cos(θ + 60◦) cos(θ + 30◦) = k
    is equivalent to no values of
    θ that satisfy √3 − 2 sin 2θ = k
    Therefore you are looking for the values of k OUTSIDE the range of the function
    Think of the range of values
    √3 − 2 sin 2θ can take. k is everything it CAN'T TAKE.
    Yes, I know there are no values.

    So I drew a sine wave, and when sin2θ=-1, at the minimum point, you want k outside of this value so I made k < root3 +2

    And vice versa with
    sin2θ=1

    I subbed in
    sin2θ=1 and sin2θ=-1 into √3 − 2 sin 2θ
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    (Original post by Peanut247)
    Yes, I know there are no values.

    So I drew a sine wave, and when sin2θ=-1, at the minimum point, you want k outside of this value so I made k < root3 +2

    And vice versa with
    sin2θ=1

    I subbed in
    sin2θ=1 and sin2θ=-1 into √3 − 2 sin 2θ
    good. that wasn't too bad, was it?
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    (Original post by CancerousProblem)
    good. that wasn't too bad, was it?
    ...but this isn't the answer on the mark scheme.

    k > root3 +2 and k<root3 -2 is correct, I have it the other way around.

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    (Original post by Peanut247)
    ...but this isn't the answer on the mark scheme.

    k > root3 +2 and k<root3 -2 is correct, I have it the other way around.

    um wait wtf r u doing
    the minimum isn't root3 + 2
    That's the MAXIMUM
    the minimum is root3 - 2, it's the SMALLEST point
    root3 + 2 is greater than root3 -2

    The MAXIMUM on the Sin function will be the MINIMUM on the NEGATIVE SIN FUNCTION
    I think that's what you're getting confused by
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    (Original post by CancerousProblem)
    um wait wtf r u doing
    the minimum isn't root3 + 2
    That's the MAXIMUM
    the minimum is root3 - 2, it's the SMALLEST point
    root3 + 2 is greater than root3 -2

    The MAXIMUM on the Sin function will be the MINIMUM on the NEGATIVE SIN FUNCTION
    I think that's what you're getting confused by
    So I should be drawing out the negative sine function not the positive one?
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    (Original post by Peanut247)
    Don't you factor in that negative when you sub -1 and +1 for sinθ into the equation?
    No. I don't really see an interpretation under which that makes sense, to be honest :P draw a little diagram of -2 \sin(2\theta).
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    (Original post by Peanut247)
    So I should be drawing out the negative sine function not the positive one?
    Well the function in the question was a negative sin function, not a positive one

    Draw a NEGATIVE sin function. Multiply the heights by 2. Now move it up by root3.
    That's the function that you have in the question.
 
 
 
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