Question

помогите пожалуйста !!​

Answer 1

C)

$tg\left (4x + \dfrac{\pi}{4}\right ) = 1\\\\\\4x + \dfrac{\pi}{4} = arctg1 + \pi k\\\\\\4x + \dfrac{\pi}{4} = \dfrac{\pi}{4} + \pi k\\\\\\4x = \dfrac{\pi}{4} + \pi k - \dfrac{\pi}{4}\\\\\\4x = \pi k\\\\\\\boxed{x = \dfrac{\pi k}{4}}\ ,\ k\in\mathbb{Z}$

А)

$\cos^2x = \sin x + 1\\\\\cos^2x - \sin x -1 = 0\\\\1 - \sin^2x - \sin x - 1 = 0\\\\-\sin^2x - \sin x = 0\\\\-\sin x(\sin x + 1) = 0\ \ \ \ \ \Big| \cdot (-1)\\\\\sin x(\sin x +1) = 0$

Произведение равно нулю, когда хотя бы один из множителей равен нулю.

$\left[\begin{gathered}\sin x = 0\\\\\sin x + 1 = 0\\\end{gathered}\ \ \ \Leftrightarrow\ \left[\begin{gathered}x = \pi k\\\\\sin x = -1\\\end{gathered}\ \ \ \Leftrightarrow\ \left[\begin{gathered}x = \pi k\\\\x = -\dfrac{\pi}{2} + 2\pi k\\\end{gathered}\ \ \ ,\ k\in\mathbb{Z}$

B)

$\sin 2x + \cos 2x - \cos^2x = 0\\\\2\sin x\cos x + \cos^2x - \sin^2x - \cos^2x = 0\\\\2\sin x\cos x - \sin^2x = 0\\\\\sin x(2\cos x -\sin x) = 0$

Произведение равно нулю, когда хотя бы один из множителей равен нулю.

$\left[\begin{gathered}\sin x = 0\\\\2\cos x -\sin x = 0\end{gathered}\ \ \ \Leftrightarrow\ \left[\begin{gathered}x = \pi k\\\\2\cos x = \sin x\ \ \ \Big| :\cos x\end{gathered}\ \ \ \Leftrightarrow\ \left[\begin{gathered}x = \pi k\\\\tgx = 2\\\end{gathered}\ \Leftrightarrow\ \left[\begin{gathered}x = \pi k\\\\x = arctg2 + \pi k\end{gathered}$