Question

дам 50 баллов! срочно! задание на фото

Answer 1

$4\cos\left (\dfrac{x}{2} + \dfrac{\pi}{8}\right ) + \sqrt{12} = 0\\\\\\4\cos\left (\dfrac{x}{2} + \dfrac{\pi}{8}\right ) = -\sqrt{12}\ \ \ \Big| :4\\\\\\\cos\left (\dfrac{x}{2} + \dfrac{\pi}{8}\right ) = -\dfrac{\sqrt{12}}{4}\\\\\\\cos\left (\dfrac{x}{2} + \dfrac{\pi}{8}\right ) = -\dfrac{2\sqrt{3}}{4}\\\\\\\cos\left (\dfrac{x}{2} + \dfrac{\pi}{8}\right ) = -\dfrac{\sqrt{3}}{2}$

$\left[\begin{gathered}\dfrac{x}{2} + \dfrac{\pi}{8} = \dfrac{5\pi}{6} + 2\pi k\\\\\dfrac{x}{2} + \dfrac{\pi}{8} = -\dfrac{5\pi}{6} + 2\pi k\\\end{gathered}\ \ \Leftrightarrow\ \left[\begin{gathered}\dfrac{x}{2} = \dfrac{17\pi}{24} + 2\pi k\\\\\dfrac{x}{2} = -\frac{23\pi}{24} + 2\pi k\\\end{gathered}\ \ \Leftrightarrow\ \left[\begin{gathered}x = \frac{17\pi}{12} + 4\pi k\\\\x = -\frac{23\pi}{12} + 4\pi k\end{gathered}\ \ ,\ k\in\mathbb{Z}$