Question

Решите функции 1) sin2x=sqrt2 cos(pi/2 + x) 2) 4cos (4x-1)+12sin^2 (4x-1) =11

Answer 1

$1)sin2x=\sqrt2* cos( \frac{\pi}{2} + x)\\ 2sinxcosx=\sqrt2*(-sinx)\\ 2sinxcosx+\sqrt2*sinx=0\\ 2sinx(cosx+ \frac{ \sqrt{2} }{2} )=0\\ cosx+ \frac{ \sqrt{2} }{2}=0\ \ ili \ \ sinx=0\\ cosx=-\frac{ \sqrt{2} }{2}\\ \\ sinx=0\\ x=\pi n, \ \ nEZ\\ \\$
$cosx=-\frac{ \sqrt{2} }{2}\\ x=б arccos(-\frac{ \sqrt{2} }{2})+2\pi n, \ nEZ\\ x=б (\pi-arccos\frac{ \sqrt{2} }{2})+2\pi n, \ nEZ\\ x=б (\pi-\frac{ \pi}{4})+2\pi n, \ nEZ\\ x=б \frac{ 3\pi}{4}+2\pi n, \ nEZ\\$