Question

6sin²x-5sinx-4=0 найти корни на отрезке [-7п/2; -3п/2]

Answer 1

$6sin^2x-5sinx-4=0\; ,\; \; \; [- \frac{7\pi }{2} ;- \frac{3\pi}{2} ]\\\\D=25+4\cdot 6\cdot 4=121\\\\(sinx)_1= \frac{5-11}{12}=- \frac{1}{2}\quad ili\quad (sinx)_2= \frac{5+11}{12}= \frac{4}{3} \\\\a)\; \; sinx=-\frac{1}{2}\; \; \; \to \; \; \; x=(-1)^{n}\cdot (-\frac{\pi}{6})+\pi n\; ,\; n\in Z \\\\x=(-1)^{n+1}\cdot \frac{\pi}{6}+\pi n\; ,\; n\in Z\\\\b)\; \; sinx=\frac{4}{3}\ \textgreater \ 1\; \; net\; \; reshenij,\; t.k.\; \; |sinx| \leq 1\\\\2)\quad x\in [-\frac{7\pi}{2};-\frac{3\pi}{2}]:\\\\x=-2\pi -\frac{\pi}{6}=-\frac{13\pi }{6}\; \; ,\; \; \; x=-2\pi -\frac{5\pi}{6}=-\frac{17\pi }{6}\\\\Otvet:\; \; x=(-1)^{n+1}\cdot \frac{\pi}{6}+\pi n,\; n\in Z\; ;\; x=-\frac{13\pi }{6}\; ,\; x=-\frac{17\pi }{6}\; .$