dream13rxs

2022-07-14

I was doing some revision and I bumped into this question.
$\mathrm{cos}\left(x\right)-\mathrm{sec}\left(x\right)=0$, ${0}^{\circ }\le x\le {360}^{\circ }$

Bruno Dixon

Expert

The equation ${\mathrm{cos}}^{2}\left(x\right)-1=0$ does not have solutions at $x={90}^{\circ },{270}^{\circ }$. This is because $\mathrm{cos}\left({90}^{\circ }\right)=0$ and $\mathrm{cos}\left({270}^{\circ }\right)=0$. Because of the stated trigonometric identity, it follows that $x={90}^{\circ },{270}^{\circ }$ are not solutions of ${\mathrm{sin}}^{2}\left(x\right)=0$

Logan Wyatt

Expert

Starting from $\frac{co{s}^{2}\left(x\right)-1}{cos\left(x\right)}=0$ which implies $\frac{-si{n}^{2}\left(x\right)}{cos\left(x\right)}=tan\left(x\right).sin\left(x\right)=0$ so these are 0 at 0,180,360 . Thus the answer is 0,180,360