Question

Answer 1

ĐK: `cosx \ne 0 <=> x \ne π/2+kπ`

`tanx-\sqrt3=1/(cosx)`

`<=> (sinx)/(cosx)-\sqrt3=1/(cosx)`

`<=>sinx-\sqrt3 cosx=1`

`<=> 1/2 sinx -\sqrt3/2 cosx =1/2`

`<=> sinx .cos π/3 - cosx . sin π/3 =1/2`

`<=> sin(x-π/3)=sin π/6`

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\text{π}}{3}=\dfrac{\text{π}}{6}+k2\text{π}\\x-\dfrac{\text{π}}{3}=\text{π}-\dfrac{\text{π}}{6}+k2\text{π}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\text{π}}{2}+k2\text{π}\left(L\right)\\x=\dfrac{7\text{π}}{6}+k2\text{π}\end{matrix}\right.\left(k\in Z\right)\)

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