Amath

4 cos x - 4 sin x + k sin x
= 4 cos x + (k - 4) sin x
= R cos ( x - t)
= R cos x cos t + R sin x sin t.
so R cos t = 4.................(1)
and R sin t = k - 4.............(2)
so R = sqrt[ 4^2 + (k - 4)^2]
so the equation is transformed to
cos ( x - t) = 5/sqrt[4^2 + (k - 4)^2]
if the equation has no solution 5/sqrt[ 4^2 + (k - 4)^2] > 1 because
cos ( x - t) cannot be greater than 1.
That is
25 > 4^2 + (k -4)^2
9 > ( k - 4)^2
0 > k^2 - 8k + 16 - 9
0 > k ^2 - 8k + 7
0 > ( k - 1)(k -7)
so 1 < k < 7.

問題應該是
IF the equation 4(cos ^2x -sin^2x)+ksin x=5 has no solution ,find the range of values of k?

2009-03-07 17:23:26 補充：
4(cos ^2x -sin^2x)+ksin x=5

4(1-2sin^2x)+ksinx=5

8sin^2x-ksinx+1=0

&becaus; the equation has no solution

∴Δ<0

(-ksinx)^2-4(8sin^2x)(1)<0

k^2sin^2x-32sin2x<0

(k^2-32)sin^2x<0

k^2-32<0

k^2<32

∴-4√2<4√2