Find the general solution for given trigonometric equation

Question

Find the general solution of:

$$ \sin^2x \cos^2x+\sin x \cos x-1=0 $$

The options are given in the form of $ \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 $ \sin2x $, through which I got

$$ \sin^22x+2\sin2x-4=0 $$

I have got no idea how to proceed further.

Answer 1

Now, $$(\sin2x+1)^2=5,$$ which is impossible because $$0\leq (\sin2x+1)^2\leq4.$$

Answer 2

Hint

First set $$u\triangleq \sin x\cos x$$to obtain $$u^2+u-1=0$$then solve for $x$ from $u$.

Answer 3

I'm sorry, but this quadratic equation has $-1\pm\sqrt 5$ 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.