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

ОДЗ:

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

С учетом ОДЗ:

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

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

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

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