\( \frac{\text { Example } 6}{\text { Solve } \cos 3 y \cos 2 y+\sin 3 y \sin 2 y=0 \text { for } 0 \leqslant y<2 \pi} \)

To solve the equation: \[ \cos 3y \cos 2y + \sin 3y \sin 2y = 0 \quad \text{for} \quad 0 \leq y < 2\pi \] we can utilize trigonometric identities to simplify the equation. ### Step 1: Apply the Cosine Difference Identity Recall the cosine difference identity: \[ \cos A \cos B + \sin A \sin B = \cos(A - B) \] Applying this to the left-hand side of the equation: \[ \cos 3y \cos 2y + \sin 3y \sin 2y = \cos(3y - 2y) = \cos y \] So, the equation simplifies to: \[ \cos y = 0 \] ### Step 2: Solve for \( y \) We need to find all angles \( y \) in the interval \( 0 \leq y < 2\pi \) where \( \cos y = 0 \). The cosine function equals zero at: \[ y = \frac{\pi}{2} \quad \text{and} \quad y = \frac{3\pi}{2} \] ### Conclusion The solutions to the equation within the specified interval are: \[ y = \frac{\pi}{2} \quad \text{and} \quad y = \frac{3\pi}{2} \] **Final Answer:** All real numbers y with y = π⁄2 and y = 3 π⁄2 in the interval from 0 to 2π.