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cos^2 x -2 cosx -3 =0

1)\begin{lgathered}cos^{2}x+2cosx-3=0\\ cosx=t\\ t^{2}+2t-3=0\\ D=4+12=16\\ t1=1\\ t2=-3\\ cosx=1\\ x=2\pi k\end{lgathered}cos2x+2cosx−3=0cosx=tt2+2t−3=0D=4+12=16t1=1t2=−3cosx=1x=2πk​2)\begin{lgathered}2tg^{2}x+3tgx-2 = 0\\ tgx=t\\ 2t^{2}+3t-2=0\\ D = 9+16=25\\ t1 = \frac{-3+5}{4} = \frac{1}{2}\\ t2 = -2\\ tgx=\frac{1}{2}\\ x=arctg\frac{1}{2}+\pi k\\ tgx=-2\\ x=-arctg2+\pi k\end{lgathered}2tg2x+3tgx−2=0tgx=t2t2+3t−2=0D=9+16=25t1=4−3+5​=21​t2=−2tgx=21​x=arctg21​+πktgx=−2x=−arctg2+πk​ )