We can use the trig identity:
[tex]1+\tan ^2\alpha=\sec ^2\alpha[/tex]
Then we rewrite and solve for tan:
[tex]\begin{gathered} 1+\tan ^2x=6-4 \\ \tan ^2x=2-1 \\ \tan ^2x=1 \end{gathered}[/tex]
Now we can apply square root on both sides:
[tex]|\tan x|=1[/tex]
(We use the absolute value because tan^2 only can give possitive results)
Finally, we can use the identity:
[tex]\tan \alpha=\frac{\sin \alpha}{\cos \alpha}[/tex]
Thus:
[tex]|\frac{\sin x}{\cos x}|=1[/tex]
We are looking for values of sine and cosine that are equal in absolute value. We know that this happens for the first time in pi/4 and happens every pi/2 from there.
Thus, the solutions are:
[tex]x=\frac{\pi}{4},\frac{3}{4}\pi,\frac{5}{4}\pi,\frac{7}{4}\pi[/tex]
