𝑺𝒐𝒍𝒗𝒆,𝒈𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕 𝟎≤𝒙<𝟐𝝅

$\sin^2 x - \sin x = 2\\ \sin^2 x - \sin x - 2 = 0 \\\text{Let a = sin x}\\ a^2 - a - 2 = 0\\ (a-2)(a+1) = 0\\ (\sin x - 2)(\sin x + 1) = 0\\ \sin x = 2\text{(rejected) or }\sin x = -1\\ \sin x = -1\\ x = \dfrac{3\pi}{2} \text{rad}$

$\cos^2 x + \cos x = \sin^2x\\ \cos^2x + \cos x = 1 - \cos^2x\\ 2\cos^2x + \cos x - 1 = 0\\ \text{Let a = cos x this time.}\\ 2a^2 + a - 1 = 0\\ (2a - 1)(a + 1) = 0\\ \cos x = \dfrac{1}{2} \text{ or }\cos x = -1\\ x = \dfrac{\pi}{3}\text{ rad} \text{ or }x = \pi \text{ rad}$

Not sure how to prove it, but just by looking at it 'a' is 3/2 pi

Nicely done, Max !  Thanx !