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--5pi; -4pi]

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Ответ:

Polina28548

03.07.2020 17:58

1)
$\sqrt2\cdot\sin(\frac{7\pi}{2}-x)\sin x=\cos x;\\
\sqrt2\left(\sin\frac{7\pi}{2}\cos x-\sin x\cos\frac{7\pi}{2}\right)\cdot\sin x=\cos x;\\
\sin\frac{7\pi}{2}=\sin(\frac{7\pi}{2}-2\pi)=\frac{7\pi-4pi}{2}=\sin\frac{\3\pi}{2}=-1;\\
\cos{\frac{7\pi}{2}}=\cos{\frac{3\pi}{2}}=0;\\
\sqrt2\cdot(-1)\cdot\cos x\sin x=\cos x;\\$
$a)\cos x=0;\ \ \ x=\frac{\pi}{2}+\pi n, n\in Z;\\
b)\cos x\neq 0;\\
-\sqrt2\sin x=1;\\
\sin x=-\frac{\sqrt2}{2};\\
x=(-1)^k\arcsin(-\frac{\sqrt2}{2})+\pi k=(-1)^{k+1}\arcsin{\frac{\sqrt2}{2}}+\pi k=\\
=(-1)^{k+1}\cdot\frac{\pi}{4}+\pi k, k\in Z;\\
 \left \{ {{x=\frac{\pi}{2}+\pi n;} \atop {x=(-1)^{k+1}\cdot\frac{\pi}{4}+\pi k}} \right.\ \ \ n,k\in Z$

2)
$\cos^2\frac{x}{2}-\sin^2\frac{x}{2}=\sin(\frac{\pi}{2}-2x);\\
\cos x=\sin\frac{\pi}{2}\cos2x-\cos\frac{\pi}{2}\sin2x;\\
\sin\frac{\pi}{2}=1;\ \ \ \cos\frac{\pi}{2}=0;\\
\cos x=\cos2x;\\
\cos x=2\cos^2x-1;\\
2\cos^2x-\cos x-1=0;\\
\cos x=t, \ \ \ -1\leq x\leq1;\\
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=2\pi n, n\in Z;\\
t_2=\frac{1-3}{4}=\frac{-2}{4}=-\frac{1}{2};\\
\cos x=-\frac{1}{2};\\
x=\pm\arccos(-\frac{1}{2})+2\pi k=\pm\frac{2\pi}{3}+2\pi k;\\
$
$
 \left[{{x=2\pi n, } \atop {x=\pm\frac{2\pi}{3}+2\pi k;}} \right. n,k\in Z;\\$

3)
$\cos2x=1-\cos(\frac{\pi}{2}-x);\ \ -5\leq x\leq-4 \ \ \ \ x\in[-5\pi;-4\pi]\\
\cos2x=1-\cos\frac{\pi}{2}\cos x+\sin x\sin\frac\pi2;\\
\cos2x=1+\sin x;\\
1-2\sin^2x=1+\sin x;\\
\sin^2x+2\sin x=0;\\
\sin x(2\sin x+1)=0;\\
a)\sin x=0;\ \ x=\pi n, n\in Z;\\
b) \sinx=-\frac{1}{2};\\
x=(-1)^{k+1}\arcsin\frac{1}{2}+\pi k=(-1)^{k+1}\frac{\pi}{6}+\pi k, k\in Z
\\
\\
 \left[{{x=\pi,n} \atop {x=(-1)^{k+1}\frac{\pi}{6}+\pi k}} \right. n,k\in Z;\\
$
$n=-5:x=-5\pi;\\
 n=-4:x=-4\pi;\\
k=-5: (-1)^{-5+1}\frac{\pi}{4}-5\pi=\frac{\pi}{4}-5\pi=\frac{(1-20)\pi}{4}=-\frac{-19\pi}{4}=-4,75\pi=\\
=-4\frac{3}{4};\\
k=-4:\ \ (-1)^{-4+1}\frac{\pi}{4}-4\pi=-\frac{\pi}{4}-4\pi=-4\frac{1}{4}\pi;\\
x=-5\pi;\ \ -4\frac{3}{4}\pi;\ \ -4\frac{1}{3};\ \ -4\pi$