If [tex]\cos x=\tan x[/tex], then
[tex]\cos x=\dfrac{\sin x}{\cos x}\implies\cos^2x=\sin x[/tex]
for [tex]\cos x\neq0[/tex]. This is definitely the case, since [tex]\tan x[/tex] is undefined if [tex]\cos x=0[/tex]. Now, [tex]\cos^2x=1-\sin^2x[/tex], so we have
[tex]1-\sin^2x=\sin x\implies\sin^2x+\sin x-1=0\implies\sin x=\dfrac{-1\pm\sqrt5}2[/tex]
or [tex]x\approx-1.618[/tex] and [tex]x\approx0.618[/tex]. One of these solutions is larger than 1 in absolute value, but [tex]|\sin x|\le1[/tex], so we omit that solution, leaving us with
[tex]\sin x=\dfrac{\sqrt5-1}2\implies x=\arcsin\left(\dfrac{\sqrt5-1}2\right)+2n\pi[/tex]
for integers [tex]n[/tex], which follows from the fact that [tex]\sin x[/tex] has period [tex]2\pi[/tex].
