Question

Ребят нужна с этими заданиями.От что у меня есть (55)

Answer 1

$18. \ \sin 2x = \sin \left(\dfrac{\pi}{2} + x \right)$

$2\sin x \cos x = \cos x$

$2\sin x \cos x - \cos x = 0$

$\cos x (2\sin x - 1) = 0$

$\left[\begin{array}{ccc}\cos x = 0, \ \ \ \ \ \ \ \ \ (1) \\2\sin x - 1 = 0 \ \ \ (2) \\\end{array}\right$

$(1) \ \cos x = 0$

$x = \dfrac{\pi}{2} + \pi n, \ n \in Z$

$(2) \ 2\sin x - 1 = 0$

$2\sin x = 1$

$\sin x = \dfrac{1}{2}$

$x = (-1)^{k} \arcsin \dfrac{1}{2}+ \pi k, \ k \in Z$

$x = (-1)^{k}\dfrac{\pi}{6} + \pi k, \ k \in Z$

ответ: $x = \dfrac{\pi}{2} + \pi n, \ x = (-1)^{k}\dfrac{\pi}{6} + \pi k, \ n \in Z, \ k \in Z$

$19. \ \displaystyle \lim_{x \to 2} \dfrac{\sqrt{x} - \sqrt{2}}{x - 2} = \left\{\dfrac{0}{0} \right\} = \lim_{x \to 2} \dfrac{\left(\sqrt{x} - \sqrt{2} \right)\left(\sqrt{x} + \sqrt{2} \right)}{(x - 2)\left(\sqrt{x} + \sqrt{2} \right)} =$

$= \displaystyle \lim_{x \to 2} \dfrac{x - 2}{(x - 2)\left(\sqrt{x} + \sqrt{2} \right)} = \lim_{x \to 2} \dfrac{1}{\sqrt{x} + \sqrt{2}} = \dfrac{1}{\sqrt{2} + \sqrt{2}} = \dfrac{1}{2\sqrt{2}} = \dfrac{\sqrt{2}}{4}$

ответ: $\dfrac{\sqrt{2}}{4}$

$20. \ y = \dfrac{e^{x/2} + \ln \cos x}{3x}$

$y' = \dfrac{(e^{x/2} +\ln\cos x)' \cdot 3x - (e^{x/2} +\ln\cos x)(3x)'}{(3x)^{2}} =$

$=\dfrac{\left( \dfrac{1}{2} e^{x/2} + \dfrac{1}{\cos x} \cdot (-\sin x)\right) \cdot 3x - (e^{x/2} + \ln \cos x) \cdot 3}{9x^{2}} =$

$= \dfrac{\dfrac{3}{2} xe^{x/2} - 3x \, \text{tg} \, x - 3e^{x/2} - 3\ln \cos x}{9x^{2}}$

ответ: $y' = \dfrac{\dfrac{3}{2} xe^{x/2} - 3x \, \text{tg} \, x - 3e^{x/2} - 3\ln \cos x}{9x^{2}}$