(3). \( \sin 4 x-\cos ^{4} x=\sin ^{2} x-\cos ^{2} x \)

The equation given is \[ \sin 4x - \cos^4 x = \sin^2 x - \cos^2 x. \] To solve this equation, we can start by simplifying both sides step by step. ### Step 1: Rewrite \(\sin 4x\) Using the double angle formula, we know that: \[ \sin 4x = 2 \sin 2x \cos 2x. \] And we can further express \(\sin 2x\) as: \[ \sin 2x = 2 \sin x \cos x. \] Thus, \[ \sin 4x = 2(2 \sin x \cos x) \cos 2x = 4 \sin x \cos x \cos 2x. \] ### Step 2: Rewrite \(\cos^4 x\) We can express \(\cos^4 x\) as: \[ \cos^4 x = (\cos^2 x)^2. \] ### Step 3: Rewrite \(\sin^2 x - \cos^2 x\) This can be rewritten using the identity: \[ \sin^2 x - \cos^2 x = -\cos 2x. \] ### Step 4: Substitute back into the equation Now substituting these back into the original equation gives us: \[ 4 \sin x \cos x \cos 2x - \cos^4 x = -\cos 2x. \] ### Step 5: Rearranging the equation Rearranging the equation leads to: \[ 4 \sin x \cos x \cos 2x + \cos 2x = \cos^4 x. \] Factoring out \(\cos 2x\) from the left side: \[ \cos 2x (4 \sin x \cos x + 1) = \cos^4 x. \] ### Step 6: Solve for \(\cos 2x\) Now we can analyze the equation. We can set \(\cos 2x = 0\) or solve for the other factor: 1. **Case 1:** \(\cos 2x = 0\) This gives us: \[ 2x = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad x = \frac{\pi}{4} + \frac{n\pi}{2}, \quad n \in \mathbb{Z}. \] 2. **Case 2:** \(4 \sin x \cos x + 1 = \frac{\cos^4 x}{\cos 2x}\) This case is more complex and may require numerical methods or further algebraic manipulation. ### Conclusion The solutions from Case 1 are: \[ x = \frac{\pi}{4} + \frac{n\pi}{2}, \quad n \in \mathbb{Z}. \] For Case 2, further analysis or numerical methods may be needed to find specific solutions. Would you like to proceed with numerical methods for Case 2 or focus on the solutions from Case 1?