Решите уравнение: 2cosx+sin2x=2cos3x2\cos x + \sin^2x = 2\cos^3x2cosx+sin2x=2cos3x. Показать решение Решение: Выразим sin2x\sin^2xsin2x через cosx\cos xcosx: 2cosx+1−cos2x=2cos3x2\cos x + 1 - \cos^2x = 2\cos^3x2cosx+1−cos2x=2cos3x 2cos3x+cos2x−2cosx−1=02\cos^3x + \cos^2x - 2\cos x - 1 = 02cos3x+cos2x−2cosx−1=0 Сделаем замену t=cosxt = \cos xt=cosx: 2t3+t2−2t−1=0⇔2t(t2−1)+t2−1=0⇔2t^3 + t^2 - 2t - 1 = 0 \Leftrightarrow 2t(t^2 - 1) + t^2 - 1 = 0 \Leftrightarrow2t3+t2−2t−1=0⇔2t(t2−1)+t2−1=0⇔ ⇔(t−1)(t+1)(2t+1)=0\Leftrightarrow (t - 1)(t + 1)(2t + 1) = 0⇔(t−1)(t+1)(2t+1)=0 t=−1илиt=1илиt=−12t = -1 \quad \text{или} \quad t = 1 \quad \text{или} \quad t = -\frac{1}{2}t=−1илиt=1илиt=−21 cosx=−1илиcosx=1илиcosx=−12\cos x = -1 \quad \text{или} \quad \cos x = 1 \quad \text{или} \quad \cos x = -\frac{1}{2}cosx=−1илиcosx=1илиcosx=−21 x=πk,x=±2π3+2πn,k,n,∈Zx = \pi k, \quad x = \pm\frac{2\pi}{3} + 2\pi n, \quad k, n, \in \mathbb{Z}x=πk,x=±32π+2πn,k,n,∈Z