The number of distinct solutions of the equation $\frac{5}{4}\cos^2 2x + \cos^4 x + \sin^4 x + \cos^6 x + \sin^6 x = 2$ in the interval $[0, 2\pi]$ is

$\text{Given: } \dfrac{5}{4}\cos^2 2x + \cos^4 x + \sin^4 x + \cos^6 x + \sin^6 x = 2$ $x \in [0, 2\pi]$ $\text{To find: number of distinct solutions in } [0, 2\pi]$ $1.\ \text{simplify } \cos^4 x + \sin^4 x \text{ and } \cos^6 x + \sin^6 x$ $2.\ \text{reduce the equation to a single trig term}$ $3.\ \text{solve and count in } [0, 2\pi]$ $\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x\cos^2 x$ $= 1 - 2\sin^2 x\cos^2 x$ $\sin 2x = 2\sin x\cos x \ \Rightarrow \ 2\sin^2 x\cos^2 x = \dfrac{1}{2}\sin^2 2x$ $\cos^4 x + \sin^4 x = 1 - \dfrac{1}{2}\sin^2 2x$ $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)^3 - 3\sin^2 x\cos^2 x(\cos^2 x + \sin^2 x)$ $= 1 - 3\sin^2 x\cos^2 x$ $\cos^6 x + \sin^6 x = 1 - \dfrac{3}{4}\sin^2 2x$ $LHS = \dfrac{5}{4}\cos^2 2x + \left(1 - \dfrac{1}{2}\sin^2 2x\right) + \left(1 - \dfrac{3}{4}\sin^2 2x\right)$ $LHS = \dfrac{5}{4}\cos^2 2x + 2 - \dfrac{5}{4}\sin^2 2x$ $LHS = \dfrac{5}{4}(\cos^2 2x - \sin^2 2x) + 2$ $\cos^2 2x - \sin^2 2x = \cos 4x$ $LHS = \dfrac{5}{4}\cos 4x + 2$ $\dfrac{5}{4}\cos 4x + 2 = 2$ $\cos 4x = 0$ $\cos\theta = 0 \Rightarrow \theta = (2n+1)\dfrac{\pi}{2}$ $4x = (2n+1)\dfrac{\pi}{2}$ $x = (2n+1)\dfrac{\pi}{8}, \quad n \in \mathbb{Z}$ $x = \dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{5\pi}{8}, \dfrac{7\pi}{8}$ $x = \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}$ $n = 0,1,\ldots,7$ $\boxed{\text{Number of distinct solutions} = 8}$