If $\cos\theta + \cos^{3}\theta = \sin^{2}\theta$, then $\sin^{6}\theta - 4\sin^{4}\theta + 8\sin^{2}\theta$ is equal to …

Given $\cos\theta + \cos^{3}\theta = \sin^{2}\theta$. Now, $\sin^{2}\theta = 1 - \cos^{2}\theta$. So, $1 - \cos^{2}\theta = \cos\theta + \cos^{3}\theta \Rightarrow \cos^{3}\theta + \cos^{2}\theta + \cos\theta - 1 = 0$. Let $\cos\theta = 1$, then $\sin\theta = 0$. Substituting $\sin\theta = 0$ gives $\sin^{6}\theta - 4\sin^{4}\theta + 8\sin^{2}\theta = 0 - 0 + 0 = 0$. But for $\cos\theta = \frac{1}{2}$, $\sin^{2}\theta = \frac{3}{4}$ gives value $4$.