数学难题。求大神解答

取仿蚂值范围稿枝为[-1，键大敏15/7]

x∈[0,2π]

f(x) =(4sinx-1)/(3cosx-4)

f'(x)

=[ (3cosx-4)(4cosx)- (4sinx-1)(-3sinx) ]/(3cosx-4)^2

=[ 12(cosx)^2 -16cosx +12(sinx)^2 -3sinx]/(3cosx-4)^2

=[ 12 -16cosx -3sinx]/(3cosx-4)^2

f'(x)=0

12 -16cosx -3sinx =0

3sinx +16cosx = 12

3tanx +16 = 12secx

(3tanx +16)^2 = 144(secx)^2

135(tanx)^2 -96tanx -112 =0

(3tanx -4)(45tanx +28)=0

tanx=4/3 or -28/45 (rej)

case 1: x 在第1象限

cosx = 3/5 , sinx=4/5

f(x) 

=(4sinx-1)/(3cosx-4)

=(16/5 -1)/(9/5 -4)

=(16-5)/(9-20)

=-1 (min)

case 2: x 在第3象限

cosx = -3/5 , sinx=-4/5

f(x) 

=(4sinx-1)/(3cosx-4)

=(-16/5 -1)/(-9/5 -4)

=(-16-5)/(-9-20)

=21/29 (max)

ie

x∈[0,2π]

f(x) =(4sinx-1)/(3cosx-4)

取值范围 =[-1, 21/29]

取值尺察缺范围没橡为[-1，15/7]，详细陵辩过程如图请参考