Đáp án:
`a) x = π/16 + k(π)/4` `(k ∈ ZZ)`
`b) x = (π)/12 + k(π)/4` `(k ∈ ZZ)`
Giải thích các bước giải:
`tan 3x - cot (x + π/4) = 0`
`<=> tan 3x = cot (x + π/4)`
`<=> tan 3x = tan ((π)/4 - x)`
`<=> 3x = π/4 - x + kπ`
`<=> 4x = π/4 + kπ`
`<=> x = π/16 + k(π)/4` `(k ∈ ZZ)`
`tan (x + π/3) + cot ((π)/6 - 3x) = 0`
`<=> tan (x + π/3) = - cot((π)/6 - 3x)`
`<=> tan (x + π/3) = cot (3x - (π)/6)`
`<=> tan (x + π/3) = tan ((2π)/3 - 3x)`
`<=> x + π/3 = (2π)/3 - 3x + kπ`
`<=> 4x = (π)/3 + kπ`
`<=> x = (π)/12 + k(π)/4` `(k ∈ ZZ)`