Đáp án+Giải thích các bước giải:

a,

`tanx=-1`

ĐK: `cosx\ne0⇔x\ne \pi/2+k\pi`      `k inZ`

PT:

`⇔tanx=tan(-\pi/4)`

`⇔x=-\pi/4+k\pi` (T/M)

Vậy `S={-\pi/4+k\pi|k inZ}`

b,

`cosx=cos``\pi/3`

`⇔[(x=\pi/3+k2\pi),(x=-pi/3+k2\pi):}`    `k inZ`

Vậy `S={\pi/3+k2\pi;-pi/3+k2\pi|k inZ}`

c,

`cos3x=sin2x`

`⇔cos3x=cos(\pi/2-2x)`

`⇔[(3x=\pi/2-2x+k2\pi),(3x=2x-\pi/2+k2\pi):}`   `k inZ`

`⇔[(5x=\pi/2+k2\pi),(x=-\pi/2+k2\pi):}`

`⇔[(x=\pi/10+(k2\pi)/5),(x=-\pi/2+k2\pi):}`

Vậy `S={\pi/10+(k2\pi)/5;-\pi/2+k2\pi| k inZ}`