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?
 A: Squaring both sides is false way  because in this case you can get false roots  $$1+\sqrt { 3 } \tan { \alpha -\sec { \alpha  } =0 } \\ 1+\sqrt { 3 } \frac { \sin { \alpha  }  }{ \cos { \alpha  }  } -\frac { 1 }{ \cos { \alpha  }  } =0\\ \cos { \alpha +\sqrt { 3 } \sin { \alpha  } =1 } \\ \frac { 1 }{ 2 } \cos { \alpha +\frac { \sqrt { 3 }  }{ 2 } \sin { \alpha  } =\frac { 1 }{ 2 }  } \\ \sin { { 30 }^{ \circ  }\cos { \alpha +\cos { { 30 }^{ \circ  }\sin { \alpha  } =\frac { 1 }{ 2 }  }  }  } \\ \sin { \left( { 30 }^{ \circ  }+\alpha  \right) =\frac { 1 }{ 2 }  } \\ { 30 }^{ \circ  }+\alpha ={ 30 }^{ \circ  }+{ 360 }^{ \circ  }n,\quad \Rightarrow \quad { \alpha  }_{ 1 }={ 360 }^{ \circ  }n,\quad \quad n=0,\pm 1,\pm 2,...\\ { 30 }^{ \circ  }+\alpha ={ 180 }^{ \circ  }-{ 30 }^{ \circ  }+{ 360 }^{ \circ  }n\quad \Rightarrow \quad { \alpha  }_{ 2 }={ 120 }^{ \circ  }+{ 360 }^{ \circ  }n,\quad n=0,\pm 1,\pm 2,.\\ \\ $$
A: Squaring will introduce spurious solutions, because you also get the solutions of
$$
1+\sqrt{3}\tan\alpha=-\sec\alpha
$$
An alternative method is to recall the formulas
$$
\sin\alpha=\frac{2t}{1+t^2}
\qquad
\cos\alpha=\frac{1-t^2}{1+t^2}
\qquad
\tan\alpha=\frac{2t}{1-t^2}
$$
where $t=\tan(\alpha/2)$, which are valid provided $\alpha\ne180^\circ$. Since this value can easily be checked not to be a solution, we can do the substitutions, getting
$$
1+\sqrt{3}\frac{2t}{1-t^2}-\frac{1+t^2}{1-t^2}=0
$$
and clearing the denominators (with the condition $t\ne\pm1$),
$$
1-t^2+2t\sqrt{3}-1-t^2=0
$$
that becomes
$$
2t(t-\sqrt{3})=0
$$
Thus we get $t=0$ or $t=\sqrt{3}$. so
$$
\tan\frac{\alpha}{2}=0
\qquad\text{or}\qquad
\tan\frac{\alpha}{2}=\sqrt{3}
$$
Since we're interested in $0\le\alpha\le360^\circ$, we limit ourselves to $0\le\alpha/2\le180^\circ$.
You should be able to finish.
