Kelly Nelson

2021-12-29

Find the general solution for given trigonometric equation
${\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+\mathrm{sin}x\mathrm{cos}x-1=0$
The options are given in the form of ${\mathrm{tan}}^{-1}$, so I tried to convert the equation completely in tan but was unable to do so.
I also tried using the identity of $\mathrm{sin}2x$, through which I got
${\mathrm{sin}}^{2}2x+2\mathrm{sin}2x-4=0$
I have got no idea how to proceed further.

Bukvald5z

Expert

Now,
${\left(\mathrm{sin}2x+1\right)}^{2}=5$,
which is impossible because
$0\le {\left(\mathrm{sin}2x+1\right)}^{2}\le 4$

Joseph Fair

Expert

Hint
First set
$u\stackrel{\mathrm{△}}{=}\mathrm{sin}x\mathrm{cos}x$
to obtain
${u}^{2}+u-1=0$
then solve for x from u.

nick1337

Expert

I'm sorry, but this quadratic equation has $-1±\sqrt{5}$ as roots, and the absolute value of each of these roots is greater than 1. Therefore as a trigonometric equation in x, it has no root.