Question

Срочно, срочно, срочно!!!! Помогите плиииз!! Алгебра!! 
1) (корень из 2)*sin(7pi/2-x)*sinx=cosx
2) cos^2*(x/2)-sin^2*(x/2)=sin(pi/2-2x)
3)cos2x=1-cos(pi/2-x)------[-5pi; -4pi]

Answer 1

1)
$sqrt2cdotsin(frac{7pi}{2}-x)sin x=cos x;\ sqrt2left(sinfrac{7pi}{2}cos x-sin xcosfrac{7pi}{2}right)cdotsin x=cos x;\ sinfrac{7pi}{2}=sin(frac{7pi}{2}-2pi)=frac{7pi-4pi}{2}=sinfrac{3pi}{2}=-1;\ cos{frac{7pi}{2}}=cos{frac{3pi}{2}}=0;\ sqrt2cdot(-1)cdotcos xsin x=cos x;$
$a)cos x=0; x=frac{pi}{2}+pi n, nin Z;\ b)cos xneq 0;\ -sqrt2sin x=1;\ sin x=-frac{sqrt2}{2};\ x=(-1)^karcsin(-frac{sqrt2}{2})+pi k=(-1)^{k+1}arcsin{frac{sqrt2}{2}}+pi k=\ =(-1)^{k+1}cdotfrac{pi}{4}+pi k, kin Z;\ left { {{x=frac{pi}{2}+pi n;} atop {x=(-1)^{k+1}cdotfrac{pi}{4}+pi k}} right. n,kin Z$

2)
$cos^2frac{x}{2}-sin^2frac{x}{2}=sin(frac{pi}{2}-2x);\ cos x=sinfrac{pi}{2}cos2x-cosfrac{pi}{2}sin2x;\ sinfrac{pi}{2}=1; cosfrac{pi}{2}=0;\ cos x=cos2x;\ cos x=2cos^2x-1;\ 2cos^2x-cos x-1=0;\ cos x=t, -1leq xleq1;\ 2t^2-t-1=0;\ D=1+8=9=(pm3)^2;\ t_1=frac{1+3}{4}=frac{4}{4}=1:cos x=1; x=2pi n, nin Z;\ t_2=frac{1-3}{4}=frac{-2}{4}=-frac{1}{2};\ cos x=-frac{1}{2};\ x=pmarccos(-frac{1}{2})+2pi k=pmfrac{2pi}{3}+2pi k;$
$left[{{x=2pi n, } atop {x=pmfrac{2pi}{3}+2pi k;}} right. n,kin Z;$

3)
$cos2x=1-cos(frac{pi}{2}-x); -5leq xleq-4 xin[-5pi;-4pi]\ cos2x=1-cosfrac{pi}{2}cos x+sin xsinfracpi2;\ cos2x=1+sin x;\ 1-2sin^2x=1+sin x;\ sin^2x+2sin x=0;\ sin x(2sin x+1)=0;\ a)sin x=0; x=pi n, nin Z;\ b) sinx=-frac{1}{2};\ x=(-1)^{k+1}arcsinfrac{1}{2}+pi k=(-1)^{k+1}frac{pi}{6}+pi k, kin Z \ \ left[{{x=pi,n} atop {x=(-1)^{k+1}frac{pi}{6}+pi k}} right. n,kin Z;$
$n=-5:x=-5pi;\ n=-4:x=-4pi;\ k=-5: (-1)^{-5+1}frac{pi}{4}-5pi=frac{pi}{4}-5pi=frac{(1-20)pi}{4}=-frac{-19pi}{4}=-4,75pi=\ =-4frac{3}{4};\ k=-4: (-1)^{-4+1}frac{pi}{4}-4pi=-frac{pi}{4}-4pi=-4frac{1}{4}pi;\ x=-5pi; -4frac{3}{4}pi; -4frac{1}{3}; -4pi$

Answer 2

Спасибо большое!!)