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

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