Question

Решите уравнение: 4 sin(x-7pi/2)=3/cosx Укажите корни, принадлежащие отрезку [-13pi/2;-5pi]

Answer 1

$4\, sin(x- \frac{7\pi }{2})= \frac{3}{cosx} \; ,\; \; \; x\in [\, - \frac{13\pi }{2},-5\pi \, ]\\\\-4\, sin(\frac{7\pi }{2}-x)= \frac{1}{cosx} \; ,\; \; \; ODZ:\; \; cosx\ne 0\; ,\; x\ne \frac{\pi }{2}+\pi n,\; n\in Z\\\\\star sin(\frac{7\pi }{2}-x)=sin(2\pi +\frac{3\pi}{2}-x)=sin( \frac{3\pi }{2}-x)=-cosx\\\\4\, cosx-\frac{3}{cosx} =0\\\\ \frac{4\, cos^2x-3}{cosx}=0\quad \Rightarrow \quad \left \{ {{4\, cos^2x-3=0,} \atop {cosx\ne 0}} \right. \\\\4\cdot cos^2x=3\; \; \to \; \; cos^2x=\frac{3}{4}\; \; ,\; \; \frac{1+cos2x}{2}=\frac{3}{4}$

$cos2x=\frac{1}{2}\\\\2x=\pm \frac{\pi}{3}+2\pi k\; ,\; k\in Z\\\\x=\pm \frac{\pi}{6}+\pi k\; ,\; k\in Z\\\\x\in [-\frac{13\pi}{2},-5\pi \, ]:\; \; x=-\frac{37\pi}{6}\; ,\; -\frac{35\pi}{6}\; ,\; -\frac{31\pi }{6}.$