Question

Решите уравнение: 2sin^2x-11cosx=-4

Answer 1

$2sin^2x-11cosx=-4\\2(1-cos^2x)-11cosx+4=0\\2-2cos^2x-11cosx+4=0\\2cos^2x+11cosx-6=0|\,\boxed{cosx=a,|a|\leq 1}\\2a^2+11a-6=0\\D^2=11^2-4*2*(-6)=121+48=169=(\pm13)^2\\a_1= \frac{-11+13}{2*2}= \frac{1}{2}\\a_2= \frac{-11-13}{2*2}=-6\,\boxed{neyd. ysl.\,(|a|\leq0)}\\cosx= \frac{1}{2}\\x=\pmarccos( \frac{1}{2} )+2\pi k,k\in\mathbb{Z}\\x= \pm\frac{\pi}{3}+2\pi k,k\in \mathbb{Z}$

Answer 2

$2sin^2x-11cosx=-4\\\\2-2cos^2x-11cosx+4=0\\\\2cos^2x+11cosx-6=0\\\\D=121+4\cdot 2\cdot 6=169=13^2\; ,\; \; -1 \leq cosx \leq 1\\\\(cosx)_1= \frac{-11-13}{4}=-6\ \textless \ -1 \; \; \; net\; reshenij\\\\(cosx)_2= \frac{-11+13}{4}= \frac{1}{2} \\\\x=\pm \arccos\frac{1}{2}+2\pi n\; ,\; n\in Z\\\\\underline {x=\pm \frac{\pi}{3} +2\pi n,\; n\in Z}$