Sin 2x cos x-3sin в квадрате x=0

To solve the equation \( \sin(2x) \cos(x) - 3 \sin^2(x) = 0 \), we will use trigonometric identities and factorization. 1. Start by using the double angle identity: \[ \sin(2x) = 2 \sin(x) \cos(x) \] Substituting this into the equation gives: \[ 2 \sin(x) \cos(x) \cos(x) - 3 \sin^2(x) = 0 \] which simplifies to: \[ 2 \sin(x) \cos^2(x) - 3 \sin^2(x) = 0 \] 2. Factor out \( \sin(x) \): \[ \sin(x) (2 \cos^2(x) - 3 \sin(x)) = 0 \] 3. Setting each factor to zero gives two cases to solve: **Case 1**: \[ \sin(x) = 0 \] The solutions are: \[ x = n\pi, \quad n \in \mathbb{Z} \] **Case 2**: \[ 2 \cos^2(x) - 3 \sin(x) = 0 \] Using the identity \( \cos^2(x) = 1 - \sin^2(x) \): \[ 2(1 - \sin^2(x)) - 3 \sin(x) = 0 \] Simplifying this gives: \[ 2 - 2 \sin^2(x) - 3 \sin(x) = 0 \] Rearranging it results in: \[ 2 \sin^2(x) + 3 \sin(x) - 2 = 0 \] 4. Now we can solve this quadratic equation in \( \sin(x) \). Let \( y = \sin(x) \): \[ 2y^2 + 3y - 2 = 0 \] Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = 2, \quad b = 3, \quad c = -2 \] \[ y = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-2)}}{2 \times 2} \] \[ y = \frac{-3 \pm \sqrt{9 + 16}}{4} \] \[ y = \frac{-3 \pm 5}{4} \] This gives: \[ y_1 = \frac{2}{4} = \frac{1}{2}, \quad y_2 = \frac{-8}{4} = -2 \quad (\text{not possible since } y = \sin(x) \in [-1, 1]) \] 5. Thus, we have: \[ \sin(x) = \frac{1}{2} \] The solutions are: \[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \] 6. Combining all the solutions, we find: - From Case 1: \( x = n\pi, \quad n \in \mathbb{Z} \) - From Case 2: \( x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \) Thus, the complete solution is: \[ x = n\pi, \quad k \in \mathbb{Z}, \quad x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \]