1) \( \frac{\sin ^{4} x+\sin ^{2} x \cos ^{2} x}{1+\cos x}=1-\cos x \)

Let's break down the equation step by step. First, we can rewrite the left-hand side: \[ \sin^4 x + \sin^2 x \cos^2 x = \sin^2 x (\sin^2 x + \cos^2 x) = \sin^2 x \cdot 1 = \sin^2 x \] Thus, the left-hand side can be rewritten as: \[ \frac{\sin^2 x}{1 + \cos x} \] Now, the equation becomes: \[ \frac{\sin^2 x}{1 + \cos x} = 1 - \cos x \] Next, we need to cross-multiply to eliminate the fraction: \[ \sin^2 x = (1 - \cos x)(1 + \cos x) \] The right-hand side simplifies to: \[ (1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x \] Now we have: \[ \sin^2 x = \sin^2 x \] This is true for all \(x\) where \(\sin^2 x\) is defined. Hence, the equation is satisfied for all \(x\) except where the original equation's left-hand side is undefined, which occurs when \(1 + \cos x = 0\) or \(\cos x = -1\). This happens at odd multiples of \(\pi\). To summarize, the original equation holds true for all \(x\) except for odd multiples of \(\pi\).