Question

Вычислить:
1)4sin(-60°)-3ctg(-60°)+5cos(-30°)
2)2sin²(-π/6)×ctg(-π/4)+3cos(-π)+6cos²(-π/4)
Уравнение:
3)ctg(2x+π/8)=√3
4)2cos²x-7sinX-5=0

Answer 1

$1)4sin(-60^0)-3ctg(-60^0)+5cos(-30^0)=-4sin60^0+3ctg60^0+5cos30^0=-4*frac{sqrt3}{2}+3*frac{sqrt3}{3}+5*frac{sqrt3}{2}=-2sqrt3+sqrt3+2.5sqrt3=1,5sqrt3$
$2)2sin^2(-frac{pi}{6})*ctg(-frac{pi}{4})+3cos(-pi)+6cos^2(-frac{pi}{4})=2sin^2frac{pi}{6}*(-ctgfrac{pi}{4})+3cospi+6cos^2frac{pi}{4}=-2*frac{1}{4}-3+6*frac{3}{4}=-frac{1}{2}-3+frac{9}{2}=1$
$3)ctg(2x+frac{pi}{8})=sqrt3\2x+frac{pi}{8}=frac{pi}{6}+pi n;nin Z\2x=frac{pi}{12}+pi n;nin Z\x=frac{pi}{24}+frac{pi n}{2};nin Z$
$4)2cos^2x-7sinx-5=0\2-2sin^2x-7sinx-5=0\2sin^2x+7sinx+3=0\sinx_{1,2}=frac{-7^+_-sqrt{49-24}}{4}\sinx_{1,2}=frac{-7^+_-5}{4}\sinx_1=-3 ;sinx_2=-frac{1}{2}\. ;x_1=-frac{pi}{6}+2pi n;nin Z;\. ;x_2=frac{7pi}{6}+2pi n;nin Z$