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Question 1

Question 2

Решите задачу:

$2)\ (64^{cosx})^{sinx}=4^{\frac{3cosx}{2}};\ x\ E\ [-\pi;\frac{\pi}{2}]\\ (4^3)^{cosx*sinx}=4^{\frac{3cosx}{2}}\\ 4^{3cosx*sinx}=4^{\frac{3cosx}{2}}\\ 3cosx*sinx=\frac{3cosx}{2}\\ 6cosx*sinx=3cosx\\ 6cosx*sinx-3cosx=0\\ 3cosx(2sinx-1)=0\\ |3cosx=0\\ |2sinx-1=0\\ \\ |cosx=0\\ |sinx=\frac{1}{2}\\ \\ |x=\frac{\pi}{2}+\pi k\\ |x=\frac{\pi}{3}+2\pi k\\ |x=\frac{2\pi}{3}+2\pi k$
$x\ E\ [-\pi;\frac{\pi}{2}]\\ 1)\ -\pi \leq \frac{\pi}{2}+\pi k \leq \frac{\pi}{2}\\ -\frac{3\pi}{2} \leq \pi k \leq 0\\ -1\frac{1}{2} \leq k \leq 0\ =\ \textgreater \ \ k=0;\ k=-1\\ k=0\ =\ \textgreater \ \ x = \frac{\pi}{2}\\ k=1\ =\ \textgreater \ \ x = \frac{\pi}{2}-\pi=-\frac{\pi}{2}\\ 2)\ -\pi \leq \frac{\pi}{3}+2\pi k \leq \frac{\pi}{2}\\ -\frac{4\pi}{3} \leq 2\pi k \leq \frac{\pi}{6}\\ -\frac{2}{3} \leq k \leq \frac{1}{12}\ =\ \textgreater \ \ k = 0\\ k = 0 \ =\ \textgreater \ \ x=\frac{\pi}{3}\\$
$3)\ -\pi \leq \frac{2\pi}{3}+2\pi k \leq \frac{\pi}{2}\\ -\frac{5\pi}{3} \leq 2\pi k \leq -\frac{\pi}{6}\\ -\frac{5}{6} \leq k \leq -\frac{1}{12}\\$
$3)\ 2cos^2x = sin(\frac{3\pi}{2}+x);\ x\ E\ [-\frac{7\pi}{2};-2\pi]\\ 2cos^2x=-cosx\\ 2cos^2x+cosx=0\\ cosx(2cosx+1)=0\\ |cosx=0\\ |2cosx+1=0\\ \\ |cosx=0\\ |cosx=-\frac{1}{2}\\ \\ |x=\frac{\pi}{2}+\pi k\\ |x=+-\frac{2\pi}{3}+2\pi k\\$
$\\ x\ E\ [-\frac{7\pi}{2};-2\pi]\\ 1)\ -\frac{7\pi}{2} \leq \frac{\pi}{2}+\pi k \leq -2\pi\\ -4\pi \leq \pi k \leq -\frac{5\pi}{2}\\ -4 \leq k \leq -2\frac{1}{2}\ =\ \textgreater \ \ k=-4;\ k=-3\\ k=-4\ =\ \textgreater \ \ x=\frac{\pi}{2}-4\pi=-\frac{7\pi}{2}\\ k=-3\ =\ \textgreater \ \ x=\frac{\pi}{2}-3\pi=-\frac{5\pi}{2}\\$
$2)\ -\frac{7\pi}{2} \leq \frac{2\pi}{3}+2\pi k \leq -2\pi\\ -\frac{25\pi}{6} \leq 2\pi k \leq -\frac{8\pi}{3}\\ -\frac{25}{12}\leq k \leq -\frac{4}{3}\\ -2\frac{1}{12} \leq k \leq -1\frac{1}{3}\ =\ \textgreater \ \ k = -2\\ k=-2\ =\ \textgreater \ \ x=\frac{2\pi}{3}-4\pi =-\frac{10\pi}{3}\\$
$3)\ -\frac{7\pi}{2}\leq -\frac{2\pi}{3}+2\pi k \leq -2\pi\\ -\frac{17\pi}{6}\leq 2\pi k \leq -\frac{4\pi}{3}\\ -\frac{17}{12}\leq k \leq -\frac{2}{3}\\ -1\frac{5}{12}\leq k \leq -\frac{2}{3}\ =\ \textgreater \ \ k=-1\\ k=-1\ =\ \textgreater \ \ x=-\frac{2\pi}{3}-2\pi=-\frac{8\pi}{3}$

Question 3

где П/3,разве там не (-1)^n П/6+Пn?

Question 4

Ох, верно.. Там П/6 + Пk и 5П/6 + Пk. Ну, так удобнее просто записывать для дальнейшего анализа. Ваш ответ тоже верен. Прошу прощения, сейчас в комментариях исправлю

Question 5

*П/6 +2Пk и 5П/6+2Пk

Question 6

-П <= П/6 + 2Пk <= П/2<br>-7П/6 <= 2Пk <= П/3<br>-7/12 <= k <= 1/6 ==> k = 0
k=0, x=П/6

Question 7

-П <= 5П/6+2Пk <=П/2<br>-11П/6 <= 2Пk <= -П/3<br>-11/12 <= k <= -1/6 - решений нет