2. Prove that 2.1 \( \operatorname{Cos}(90-2 x) \tan \left(180^{\circ}+x\right)+\sin ^{2}\left(360^{\circ}-x\right)=3 \sin ^{2} x \)

To prove the equation \( \operatorname{Cos}(90^\circ - 2x) \tan(180^\circ + x) + \sin^2(360^\circ - x) = 3\sin^2 x \), let's simplify each term step by step. 1. **Simplifying \( \operatorname{Cos}(90^\circ - 2x) \)**: - We know that \( \operatorname{Cos}(90^\circ - \theta) = \sin(\theta) \), hence: \[ \operatorname{Cos}(90^\circ - 2x) = \sin(2x) \] 2. **Simplifying \( \tan(180^\circ + x) \)**: - Using the identity \( \tan(180^\circ + \theta) = \tan(\theta) \), we get: \[ \tan(180^\circ + x) = \tan(x) \] 3. **Combining the first term**: - Now substitute back into the equation: \[ \sin(2x) \tan(x) \] 4. **Using the double angle identity for sine**: - We know that \( \sin(2x) = 2 \sin(x) \cos(x) \): \[ \sin(2x) \tan(x) = 2 \sin(x) \cos(x) \cdot \frac{\sin(x)}{\cos(x)} = 2 \sin^2(x) \] 5. **Simplifying \( \sin^2(360^\circ - x) \)**: - We also have \( \sin(360^\circ - x) = -\sin(x) \) (as sine is periodic with period \( 360^\circ \)): \[ \sin^2(360^\circ - x) = \sin^2(x) \] 6. **Bringing it all together**: - Now combine everything: \[ 2 \sin^2(x) + \sin^2(x) = 3 \sin^2(x) \] Thus, we have shown that: \[ \sin(2x) \tan(180^\circ + x) + \sin^2(360^\circ - x) = 3 \sin^2 x \] This proves the given equation is valid!