Đáp án:
89. B
90. B
Giải thích các bước giải:
\(\eqalign{
& 89)\,\,2\cos 4x - 2\left( {\sqrt 3 + 1} \right)\cos 2x + 2 + \sqrt 3 = 0 \cr
& \Leftrightarrow 2\left( {2{{\cos }^2}2x - 1} \right) - 2\left( {\sqrt 3 + 1} \right)\cos 2x + 2 + \sqrt 3 = 0 \cr
& \Leftrightarrow 4{\cos ^2}2x - 2\left( {\sqrt 3 + 1} \right)\cos 2x + \sqrt 3 = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos 2x = {1 \over 2} \hfill \cr
\cos 2x = {{\sqrt 3 } \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
2x = \pm {\pi \over 3} + k2\pi \hfill \cr
2x = \pm {\pi \over 6} + k2\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = \pm {\pi \over 6} + k\pi \hfill \cr
x = \pm {\pi \over {12}} + k\pi \hfill \cr} \right. \cr
& x \in \left( {0;\pi } \right) \cr
& \Rightarrow x \in \left\{ {{\pi \over 6};{{5\pi } \over 6};{\pi \over {12}};{{11\pi } \over {12}}} \right\} \cr
& Chon\,\,B. \cr
& 90)\,\,{{\cos 2x + 3\sin x - 2} \over {6{x^2} - \pi x - {\pi ^2}}} = 0 \cr
& DK:\,\,6{x^2} - \pi x - {\pi ^2} \ne 0 \Leftrightarrow \left\{ \matrix{
x \ne - {\pi \over 3} \hfill \cr
x \ne {\pi \over 2} \hfill \cr} \right. \cr
& PT \Leftrightarrow \cos 2x + 3\sin x - 2 = 0 \cr
& \Leftrightarrow 1 - 2{\sin ^2}x + 3\sin x - 2 = 0 \cr
& \Leftrightarrow 2{\sin ^2}x - 3\sin x + 1 = 0 \cr
& \Leftrightarrow \left[ \matrix{
\sin x = 1 \hfill \cr
\sin x = {1 \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\pi \over 2} + k2\pi \hfill \cr
x = {\pi \over 6} + k2\pi \hfill \cr
x = {{5\pi } \over 6} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr
& x \in \left( {0;\pi } \right) \Rightarrow x \in \left( {{\pi \over 2};{\pi \over 6};{{5\pi } \over 6}} \right) \cr
& Ma\,\,x \ne {\pi \over 2};\,\,x \ne - {\pi \over 3} \Rightarrow x \in \left( {{\pi \over 6};{{5\pi } \over 6}} \right) \cr
& Chon\,\,B. \cr} \)
