1 câu trả lời
Đáp án:
\(\begin{array}{l} a)\,\,\left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \pi + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ b)\,\,\,x = \pi + k4\pi \,\,\,\left( {k \in Z} \right) \end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} 1)\,\,\sin x - \cos x = 1\\ \Leftrightarrow \sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = 1\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x - \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\ x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \pi + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right).\\ b)\,\,\,\cos x - \sin \frac{x}{2} + 2 = 0\\ \Leftrightarrow 1 - 2{\sin ^2}\frac{x}{2} - \sin \frac{x}{2} + 2 = 0\\ \Leftrightarrow 2{\sin ^2}\frac{x}{2} + \sin \frac{x}{2} - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin \frac{x}{2} = 1\,\,\,\,\,\left( {tm} \right)\\ \sin \frac{x}{2} = - \frac{{3\,}}{2}\,\,\,\,\left( {ktm} \right) \end{array} \right.\\ \Leftrightarrow \frac{x}{2} = \frac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \pi + k4\pi \,\,\,\left( {k \in Z} \right). \end{array}\)
