Question

А) 2cos(x+pi/3)-корень из 3=0 б) sin^2(х-pi/4)=0,75 в) tg(pi/4-2x)=1

Answer 1

$a)\; 2cos(x+\frac{\pi}{3})-\sqrt3=0\\\\cos(x+\frac{\pi}{3})=\frac{\sqrt3}{2}\\\\x+\frac{\pi}{3}=\pm arccos\frac{\sqrt3}{2}+2\pi n=\pm \frac{\pi}{6}+2\pi n\\\\x=-\frac{\pi}{3}\pm \frac{\pi}{6}+2\pi n= \left [ {{-\frac{\pi}{2}+2\pi n,\; n\in Z} \atop {-\frac{\pi}{6}+2\pi n,\; n\in Z}} \right.$

$b)\; sin^2(x-\frac{\pi}{4})=0,75\\\\ \frac{1-cos(2x-\frac{\pi}{2})}{2} =\frac{3}{4}\\\\cos(2x-\frac{\pi}{2})=cos(\frac{\pi}{2}-2x)=sin2x=-\frac{1}{2}\\\\2x=(-1)^{n}\cdot (-\frac{\pi}{6})+\pi n=(-1)^{n+1}\frac{\pi}{6}+\pi n,\; n\in Z\\\\x=(-1)^{n+1}\frac{\pi}{12}+\frac{\pi n}{2},\; n\in Z$

$c)\; tg(\frac{\pi}{4}-2x)=1\\\\\frac{\pi}{4}-2x=\frac{\pi}{4}+\pi n,\; n\in Z\\\\-2x=\pi n,\; n\in Z\\\\x=-\frac{\pi n}{2}=\frac{\pi m}{2}\; ,\; n,\, m\in Z$