Question

Tổng các nghiệm thuộc đoạn [-\(\pi\); \(\pi\)] của phương trình tan(4x + \(\dfrac{9\pi}{8}\)) + cot(2x - \(\dfrac{3\pi}{4}\))=0

Answer 1

`tan(4x+[9\pi]/8)+cot(2x-[3\pi]/4)=0` `ĐK: x \ne -\pi/8+k\pi/2,x \ne [3\pi]/32+k\pi/4,k in ZZ`

`<=>tan(4x+\pi/8)+cot(2x+\pi/4)=0`

`<=>tan (4x+\pi/8)=-cot(2x+\pi/4)`

`<=>tan(4x+\pi/8)=tan(-2x-\pi/4)`

`<=>4x+\pi/8=-2x-\pi/4+k\pi`

`<=>x=-\pi/16+k\pi/6`  `(k in ZZ)`   (t/m)

  Mà `x in [-\pi;\pi]`

  `=>-\pi <= -\pi/16+k\pi/6 <= \pi`

`<=>-5,625 <= k <= 6,375`   Mà `k in ZZ`

   `=>k={-4;-3;-2;-1;0;1;2;3;4;5;6}`

`=>T=-\pi/16 .11+[-4\pi]/6+[-3\pi]/6+[-2\pi]/6+[-\pi]/6+\pi/6+[2\pi]/6+[3\pi]/6+[4\pi]/6+[5\pi]/6+[6\pi]/6=[55\pi]/48`