Doable soviet problem

The trivial solution for the equation is $\sin{x} = \cos{x} = \pm \frac{1}{\sqrt{2}} \Longrightarrow x = 45^\circ, 225^\circ$ . But are there other solutions.

Divide the LHS and RHS by $\sin{x}-\cos{x}$ and we get:

$\dfrac {LHS}{\sin{x}-\cos{x} } = \dfrac{(\sin{x}-\cos{x})(\sin^4{x}+\cos^4{x}) + \sin{x}\cos{x}(\sin^3{x}-\cos^2{x} )}{\sin{x}-\cos{x}} \\ = \dfrac{(\sin{x}-\cos{x})(\sin^4{x}+\cos^4{x}) + \sin{x}\cos{x}[(\sin{x}-\cos{x})(\sin^2{x}+\cos^2{x} ) \sin^2{x} \cos^2{x}(\sin{x}-\cos{x})]}{\sin{x}-\cos{x}} \\ = (\sin^4{x}+2\sin^2{x} \cos^{x} + \cos^4{x}) - 2 \sin^2{x} \cos^2{x} + \sin{x}\cos{x} + \sin^2{x} \cos^2{x} \\ = 1 + \sin{x}\cos{x} - \sin^2{x} \cos^2{x}$

$\dfrac{RHS}{\sin{x}-\cos{x}} = \dfrac{1}{\sin{x}\cos{x}}$

$\Rightarrow 1 + \sin{x}\cos{x} - \sin^2{x} \cos^2{x} = \dfrac{1}{\sin{x}\cos{x}} \\ \quad \sin{x}\cos{x}(1 + \sin{x}\cos{x} - \sin^2{x} \cos^2{x}) = 1 \\ \quad \sin{x}^3\cos^3{x} - \sin^2{x}\cos^2{x} - \sin{x}\cos{x} + 1 = 0$

$\quad \Rightarrow \sin{x}\cos{x} = 1\quad$ which has no real root.

Therefore, the solutions for $x$ are $45^\circ$ and $225^\circ$ and their sum $=\boxed{270}$ .

You factor it. Then, somewhere, you should see the pythagorean identity $\sin^{2} x + \cos^{2} x$ which is equal to $1$ . Also, make sure you set the equation equal to zero by moving $\displaystyle\frac{1}{\cos x} - \frac{1}{\sin x}$ to the LHS , and I'd suggest finding a common denominator between the two.