Đáp án:

`3)` `D=RR\\{pi/4+(kpi)/2, k in ZZ}`

`4)` `D=RR\\{-pi/2+k2pi, k in ZZ}`

`5)` `D=RR\\{pi/2+k2pi, k in ZZ}`

`6)` `D=RR\\{pi+k2pi, k in ZZ}`

`7)` `D=RR\\{pi/2+kpi, k in ZZ}`

`8)` `D=RR\\{k2pi; -pi/2+k2pi, k in ZZ}`

Giải thích các bước giải:

`3)`

ĐKXĐ: `cos 2x ne 0`

`<=>2x ne pi/2+kpi`

`<=> x ne pi/4+(kpi)/2, k in ZZ`

`=>D=RR\\{pi/4+(kpi)/2, k in ZZ}`

`4)`

ĐKXĐ: `sinx+1 ne 0`

`<=> sin x ne -1`

`<=> x ne -pi/2+k2pi, k in ZZ`

`=>D=RR\\{-pi/2+k2pi, k in ZZ}`

`5)`

ĐKXĐ: `4sinx-4 ne 0`

`<=> sin x ne 1`

`<=>x ne pi/2+k2pi, k in ZZ`

`=>D=RR\\{pi/2+k2pi, k in ZZ}`

`6)`

ĐKXĐ: `cos\ x/2 ne 0`

`<=> x/2 ne pi/2+kpi`

`<=> x ne pi+k2pi, k in ZZ`

`=>D=RR\\{pi+k2pi, k in ZZ}`

`7)`

ĐKXĐ: `{(cos x ne 0),(cosx+3 ne 0):}`

`<=>{(x ne pi/2+kpi),(cos x ne -3\ "(luôn đúng)"):}`

`=>D=RR\\{pi/2+kpi, k in ZZ}`

`8)`

ĐKXĐ: `{(sin\ x/2 ne 0),(7sinx+7 ne 0):}`

`<=>{(x/2 ne kpi),(7sinx ne -7):}`

`<=>{(x ne k2pi),(sinx ne -1):}`

`<=>{(x ne k2pi),(x ne -pi/2+k2pi):}`

`=>D=RR\\{k2pi; -pi/2+k2pi, k in ZZ}`