Question

Log2(2cos^2x + 3cosx +1)= 1 промежуток [-7pi/2 ; -2pi]

Answer 1

$\log_2(2\cos^2x+3\cos x+1)=1;\ \ x\in[\frac{-7\pi}{2};-2\pi];\\ \log_2(2\cos^2x+3\cos x+1)=\log_22;\\ 2\cos^2x+3\cos x+1=2;\\ 2\cos^2x+3\cos x-1=0;\\ t=\cos x;\\ -1\leq t\leq 1;\\ 2t^2+3t-1=0;\\ D=b^2-4ac=3^2-4\cdot2\cdot(-1)=9+8=17;\\ t_1=\frac{-b-\sqrt D}{2a}=\frac{-3-\sqrt{17}}{4}<-1\\ t_2=\frac{-b+\sqrt D}{2a}=\frac{-3+\sqrt{17}}{4}<1;\\ \cos x=\frac{\sqrt{17}-3}{4};\\ x=\pm\arccos(\frac{\sqrt{17}-3}{4})+2\pi n, n\in Z;\\ n=-1; 0\leq \arccos\leq2\pi\\$