Solve for $\alpha$ in $1+\sqrt3\tan\alpha-\sec\alpha=0$ 
Solve for $\alpha$, where $0^\circ\le\alpha\le360^\circ$, in
  $$1+\sqrt3\tan\alpha-\sec\alpha=0$$

I solved this way. But the answer in my book is only $0^\circ,120^\circ,360^\circ$.

Please help me. Which answer is suitable?

Squaring both sides is false way  because in this case you can get false roots  $$1+\sqrt { 3 } \tan { \alpha -\sec { \alpha  } =0 } \\ 1+\sqrt { 3 } \frac { \sin { \alpha  }  }{ \cos { \alpha  }  } -\frac { 1 }{ \cos { \alpha  }  } =0\\ \cos { \alpha +\sqrt { 3 } \sin { \alpha  } =1 } \\ \frac { 1 }{ 2 } \cos { \alpha +\frac { \sqrt { 3 }  }{ 2 } \sin { \alpha  } =\frac { 1 }{ 2 }  } \\ \sin { { 30 }^{ \circ  }\cos { \alpha +\cos { { 30 }^{ \circ  }\sin { \alpha  } =\frac { 1 }{ 2 }  }  }  } \\ \sin { \left( { 30 }^{ \circ  }+\alpha  \right) =\frac { 1 }{ 2 }  } \\ { 30 }^{ \circ  }+\alpha ={ 30 }^{ \circ  }+{ 360 }^{ \circ  }n,\quad \Rightarrow \quad { \alpha  }_{ 1 }={ 360 }^{ \circ  }n,\quad \quad n=0,\pm 1,\pm 2,...\\ { 30 }^{ \circ  }+\alpha ={ 180 }^{ \circ  }-{ 30 }^{ \circ  }+{ 360 }^{ \circ  }n\quad \Rightarrow \quad { \alpha  }_{ 2 }={ 120 }^{ \circ  }+{ 360 }^{ \circ  }n,\quad n=0,\pm 1,\pm 2,.\\ \\ $$
