Question

If cosx = tanx, determine all possible values of sinx.

Answer 1

If $\cos x=\tan x$, then

$\cos x=\dfrac{\sin x}{\cos x}\implies\cos^2x=\sin x$

for $\cos x\neq0$. This is definitely the case, since $\tan x$ is undefined if $\cos x=0$. Now, $\cos^2x=1-\sin^2x$, so we have

$1-\sin^2x=\sin x\implies\sin^2x+\sin x-1=0\implies\sin x=\dfrac{-1\pm\sqrt5}2$

or $x\approx-1.618$ and $x\approx0.618$. One of these solutions is larger than 1 in absolute value, but $|\sin x|\le1$, so we omit that solution, leaving us with

$\sin x=\dfrac{\sqrt5-1}2\implies x=\arcsin\left(\dfrac{\sqrt5-1}2\right)+2n\pi$

for integers $n$, which follows from the fact that $\sin x$ has period $2\pi$.