Question

Решить неравенство: ½ lg81-lgx> lg2 решить уравнение: 5(sinx)2+8cosx+1=[cosx]+(cosx)2

Answer 1

ОДЗ:

$x\ \textgreater \ 0 \\ \\ \frac{1}{2} \lg81 - \lg{x}\ \textgreater \ \lg2 \\ \\ \lg(81)^{\frac{1}{2}}- \lg x \ \textgreater \ \lg2 \\ \\ \lg\sqrt{81}- \lg x \ \textgreater \ \lg2 \\ \\ \lg9- \lg x \ \textgreater \ \lg2 \\ \\ -\lg x \ \textgreater \ \lg 2 - \lg 9 \\ \\ \lg x \ \textless \ \lg 9 - \lg 2 \\ \\ \lg x \ \textless \ \lg \frac{9}{2} \\ \\ x\ \textless \ \frac{9}{2}$

С учетом ОДЗ:

$0\ \textless \ x\ \textless \ \frac{9}{2}$

$\\ \\ 5\sin^2x+8 \cos x+1=|cosx|+\cos^2x \\ \\ 5\sin^2x+8 \cos x+ \cos^2 x + \sin^2 x=|cosx|+\cos^2x \\ \\ 6\sin^2x+8 \cos x=|cosx| \\\\ 6 \cdot (1- \cos^2{x})+8 \cos x=|cosx| \\\\ 6- 6\cos^2{x}+8 \cos x=|cosx| \\\\ \\ 1) \ \cos x \ \textgreater \ 0 \ \ \ \ \ \ \ \ \ \ 2) \ \cos \ \textless \ 0 \\ 1) \ 6-6\cos^2 x + 8\cos x = \cos x \ \ \ \ 2) \ 6-6\cos^2 x + 8\cos x = -\cos x \\ \\ 1) \ -6\cos^2 x + 7\cos x + 6= 0 \ \ \ \ 2) \ -6\cos^2 x + 9\cos x + 6= 0 \\ \\ 1) \ 6\cos^2 x - 7\cos x - 6= 0 \ \ \ \ 2) \ 6\cos^2 x - 9\cos x - 6= 0$

$t=\cos x \ \ [ \ 1) \ 0 \leq t \leq 1; \ \ \ 2) \ -1 \leq t \leq 0 \ ] \\ \\ 1) \ 6t^2 - 7t - 6= 0 \ \ \ \ 2) \ 6t^2 - 9t- 6= 0 \ \Rightarrow \ 2t^2 -3t-2=0 \\ \\ 1) \ t_{1,2}=\frac{7 \pm \sqrt{49 + 144}}{12}=\frac{7 \pm \sqrt{193}}{12}; \ \ \ 2) \ t_{3,4}=\frac{3 \pm \sqrt{9 +16}}{4}=\frac{3 \pm 5}{4} \\ \\ 1) \ t_1 =\frac{7}{12}+ \frac{\sqrt{193}}{12}\ \textgreater \ 1; \ t_2= \frac{7}{12}-\frac{\sqrt{193}}{12} \ \textless \ 0; \ \ \ 2) \ t_3=2 \ \textgreater \ 1; \ \ \ t_4=-\frac{1}{2} \\ \\ cos x= -\frac{1}{2}$

$\\ x=\pm \frac{2 \pi }{3} + 2\pi n, \ n \in Z$

Answer 2

ОДЗ x>0
1/2lg81=lg√81=lg9
lgx<lg9-lg2
lgx<lg4,5
x<4,5
x∈(0;4,5)