Homogeneous second-degree inequality reduced to linear form

Problem

Transform into linear form in sine and cosine with a suitable substitution and solve: $-\sin^{2}x-2\sin x\cos x+3\cos^{2}x-2<0.$

Solution

With $\sin^{2}x=\dfrac{1-\cos 2x}{2}$, $\cos^{2}x=\dfrac{1+\cos 2x}{2}$, $2\sin x\cos x=\sin 2x$ the left-hand side becomes $2\cos 2x-\sin 2x-1.$ Let $t=2x$: $\,2\cos t-\sin t<1$, i.e. $\sqrt5\cos(t+\varphi)<1$ with $\varphi=\arctan\tfrac12$. Hence $\cos(t+\varphi)<\dfrac{1}{\sqrt5}$, giving $t\in(0{,}6435,\ \tfrac{3\pi}{2})$ (over one period). Returning to $x=\tfrac{t}{2}$ (period $\pi$): $\boxed{\,\tfrac12\arctan\tfrac34+k\pi<x<\tfrac{3\pi}{4}+k\pi\ \ (\approx 0{,}322<x<2{,}356)\,}$