(1+sin^2 x)cosx+(1+cos^2x)sinx=1+sin2x... câu 2: sin2x+cos2x +3sinx-cosx-2 =0.

Đáp án:

\[1)\,\,\,\left[ \begin{array}{l} x = - \frac{\pi }{4} + k\pi \,\,\,\\ x = k2\pi \\ x = \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in Z} \right)\]

Giải thích các bước giải: \[\begin{array}{l} 1)\,\,\left( {1 + {{\sin }^2}x} \right)\cos x + \left( {1 + {{\cos }^2}x} \right)\sin x = 1 + \sin 2x\\ \Leftrightarrow \cos x + {\sin ^2}x\cos x + \sin x + {\cos ^2}x\sin x = 1 + 2\sin x\cos x\\ \Leftrightarrow \left( {\cos x + \sin x} \right) + \sin x\cos x\left( {\sin x + \cos x} \right) = {\left( {\sin x + \cos x} \right)^2}\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {1 + \sin x\cos x - \sin x - \cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x + \cos x = 0\,\,\,\,\,\left( 1 \right)\\ 1 + \sin x\cos x - \sin x - \cos x = 0\,\,\,\,\left( 2 \right) \end{array} \right.\\ \left( 1 \right) \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi \Leftrightarrow x = - \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right)\\ \left( 2 \right) \Leftrightarrow \sin x + \cos x - \sin x\cos x - 1 = 0\,\,\,\left( * \right)\\ Dat\,\,\,\sin x + \cos x = t\,\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right)\\ \Rightarrow {t^2} = 1 + 2\sin x\cos x\\ \Rightarrow \sin x\cos x = \frac{{{t^2} - 1}}{2}.\\ \Rightarrow \left( * \right) \Leftrightarrow t - \frac{{{t^2} - 1}}{2} - 1 = 0 \Leftrightarrow 2\left( {t - 1} \right) - \left( {t - 1} \right)\left( {t + 1} \right) = 0\\ \Leftrightarrow \left( {t - 1} \right)\left( {2 - t - 1} \right) = 0 \Leftrightarrow \left( {t - 1} \right)\left( {1 - t} \right) = 0\\ \Leftrightarrow t = 1\,\,\left( {tm} \right) \Leftrightarrow \sin x + \cos x = 1\\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 1 \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\ x + \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \frac{\pi }{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\]