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1. 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?
2. draw a sketch. it's pretty obvious why your answer can't possibly be right
3. (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?

4. 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.
5. (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 is 1, yes, but that makes the *minimum* of . (Because of the extra negative before the sin term.)
6. (Original post by Smaug123)
You're using "minimum" and "maximum" a bit sloppily, which might be the problem. The maximum of is 1, yes, but that makes the *minimum* of . (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?
7. (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
8. (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?
9. (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.

10. (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
11. (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?
12. (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 .
13. (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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