Đáp án:

Giải thích các bước giải: \[\begin{array}{l} 1.Dk:Sinx - {\mathop{\rm cosx}\nolimits} \ne 0\\ \Leftrightarrow sin(x - \frac{\pi }{4}) \ne 0 \Leftrightarrow x \ne \frac{\pi }{4} + k\pi \end{array}\] \[\begin{array}{l} 2.Vi: - 1 \le {\mathop{\rm cosx}\nolimits} \le 1\\ = > Pt\_vn\_khi:\left[ \begin{array}{l} m > 1\\ m < - 1 \end{array} \right. \end{array}\] \[\begin{array}{l} 3.Dk:x \ge 0 = > Txd:D = \left[ {0; + \infty } \right)\\ 4.\sin 2x = - \frac{1}{2}\\ = > \left[ \begin{array}{l} 2x = \frac{{ - \pi }}{6} + k2\pi \\ 2x = \frac{{7\pi }}{6} + k2\pi \end{array} \right. = > \left[ \begin{array}{l} x = \frac{{ - \pi }}{{12}} + k\pi \\ x = \frac{{7\pi }}{{12}} + k\pi \end{array} \right.\\ x \in \left( {0;\pi } \right) = > \left[ \begin{array}{l} x = \frac{{11\pi }}{{12}}\\ x = \frac{{7\pi }}{{12}} \end{array} \right. \end{array}\] \[\begin{array}{l} 5.\sin x = \frac{1}{2}\\ = > \left[ \begin{array}{l} x = \frac{\pi }{6} + k2\pi \\ x = \frac{{5\pi }}{6} + k2\pi \end{array} \right..x \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] = > x = \frac{\pi }{6}\\ 6.Dkxd:\cos x \ne 0 = > x \ne \frac{\pi }{2} + k\pi \\ 7.A\\ Vi: - 1 \le \sin x \le 1\\ 8.Dkxd:1 - \cos x \ne 0 = > x \ne k2\pi \\ 9.B\\ 10.\cos 3x = cos{12^0}\\ {12^0} = \frac{\pi }{{15}} = > 3x = \pm \frac{\pi }{{15}} + k2\pi = > x = \pm \frac{\pi }{{45}} + \frac{{k2\pi }}{3}\\ 11.\sin (x + \frac{\pi }{4}) = 1 = > x + \frac{\pi }{4} = \frac{\pi }{2} + k2\pi = > x = \frac{\pi }{4} + k2\pi \\ x \in \left[ {\pi ;3\pi } \right] = > x = \frac{{9\pi }}{4} \end{array}\]

\[\begin{array}{l} Cau\,\,1:\\ DK:\,\,\,\sin x - \cos x \ne 0\\ \Leftrightarrow \sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) \ne 0\\ \Leftrightarrow x - \frac{\pi }{4} \ne k\pi \Leftrightarrow x \ne \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right).\\ \Rightarrow chon\,\,D.\\ Cau\,\,2:\\ \cos x - m = 0\, \Leftrightarrow \cos x = m\,\,\,\left( * \right)\\ Ta\,\,co:\,\, - 1 \le \cos x \le 1\,\,\,\forall x\\ \Rightarrow pt\,\,\,\left( * \right)\,\,\,\,VN \Leftrightarrow \left[ \begin{array}{l} m > 1\\ m < - 1 \end{array} \right.\\ \Rightarrow Chon\,\,A.\\ Cau\,\,3:\\ y = \cos \sqrt x \\ DK:\,\,\,x \ge 0\\ \Rightarrow Chon\,\,\,B.\\ Cau\,\,4:\,\,\,\\ \sin 2x = - \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l} 2x = - \frac{\pi }{6} + k2\pi \\ 2x = \frac{{7\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{{12}} + k\pi \\ x = \frac{{7\pi }}{{12}} + k\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ 0 < x < \pi \\ \Rightarrow \left[ \begin{array}{l} 0 < - \frac{\pi }{{12}} + k\pi < \pi \\ 0 < \frac{{7\pi }}{{12}} + k\pi < \pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \frac{1}{{12}} < k < \frac{{13}}{{12}}\\ - \frac{7}{{12}} < k < \frac{5}{{12}} \end{array} \right. \Rightarrow \left[ \begin{array}{l} k = 1\\ k = 0 \end{array} \right.\\ \Rightarrow co\,\,2\,\,\,nghiem\,\,\,thoa\,\,man\,\,\,bai\,\,toan.\\ \Rightarrow Chon\,\,C.\\ Cau\,\,\,5\,\,\,\,tuong\,\,tu\,\,cau\,\,4.\\ \Rightarrow Chon\,\,\,B.\\ Cau\,\,6:\\ HS\,\,xd \Leftrightarrow \cos x \ne 0 \Leftrightarrow x \ne \frac{\pi }{2} + k\pi \,\,\left( {k \in Z} \right)\\ \Rightarrow Chon\,\,\,B.\\ Cau\,\,7:\\ Chon\,\,\,A\,\,\,vi\,\, - 1 \le \sin x \le 1.\\ Cau\,\,8:\\ DK:\,\,1 - \cos x \ne 0\\ \Rightarrow Chon\,\,\,A.\\ Cau\,\,9:\\ Chon\,B. \end{array}\] Các câu còn lại bạn làm tương tự nhé!!!