Решите уравнение cos22x=34\cos ^2 2x = \frac{3}{4}cos22x=43 Показать решение Решение: cos22x=34⇔cos2x=±32\cos^2 2x = \frac{3}{4} \Leftrightarrow \cos 2x = \pm\frac{\sqrt{3}}{2}cos22x=43⇔cos2x=±23 cos2x=32⇒2x=±π6+2πk,k∈Z\cos 2x = \frac{\sqrt{3}}{2} \Rightarrow 2x = \pm\frac{\pi}{6} + 2\pi k, k \in Zcos2x=23⇒2x=±6π+2πk,k∈Z cos2x=−32⇒2x=±5π6+2πk,k∈Z\cos 2x = -\frac{\sqrt{3}}{2} \Rightarrow 2x = \pm\frac{5\pi}{6} + 2\pi k, k \in Zcos2x=−23⇒2x=±65π+2πk,k∈Z 2x=π6+2πk⇒x=π12+πk,k∈Z2x = \frac{\pi}{6} + 2\pi k \Rightarrow x = \frac{\pi}{12} + \pi k, k \in Z2x=6π+2πk⇒x=12π+πk,k∈Z 2x=−π6+2πk⇒x=−π12+πk,k∈Z2x = -\frac{\pi}{6} + 2\pi k \Rightarrow x = -\frac{\pi}{12} + \pi k, k \in Z2x=−6π+2πk⇒x=−12π+πk,k∈Z 2x=5π6+2πk⇒x=5π12+πk,k∈Z2x = \frac{5\pi}{6} + 2\pi k \Rightarrow x = \frac{5\pi}{12} + \pi k, k \in Z2x=65π+2πk⇒x=125π+πk,k∈Z 2x=−5π6+2πk⇒x=−5π12+πk,k∈Z2x = -\frac{5\pi}{6} + 2\pi k \Rightarrow x = -\frac{5\pi}{12} + \pi k, k \in Z2x=−65π+2πk⇒x=−125π+πk,k∈Z
