4(sin$x^{2}$ + $\frac{1}{sinx^{2}}$) - 4(sinx-$\frac{1}{sinx}$)=7 giải giúp e với
1 câu trả lời
\[\begin{array}{l} 4\left( {{{\sin }^2}x + \frac{1}{{{{\sin }^2}x}}} \right) - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) = 7\,\,\,\,\left( {DK:\,\,\,\sin x \ne 0} \right)\\ \Leftrightarrow 4\left( {{{\sin }^2}x - 2 + \frac{1}{{{{\sin }^2}x}} + 2} \right) - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) = 7\\ \Leftrightarrow 4{\left( {\sin x - \frac{1}{{\sin x}}} \right)^2} + 8 - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) - 7 = 0\\ \Leftrightarrow 4{\left( {\sin x - \frac{1}{{\sin x}}} \right)^2} - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) + 1 = 0\\ \Leftrightarrow {\left[ {2\left( {\sin x - \frac{1}{{\sin x}}} \right) - 1} \right]^2} = 0\\ \Leftrightarrow 2\left( {\sin x - \frac{1}{{\sin x}}} \right) - 1 = 0\\ \Leftrightarrow \sin x - \frac{1}{{\sin x}} = \frac{1}{2}\\ \Leftrightarrow 2{\sin ^2}x - 2 = \sin x\\ \Leftrightarrow 2{\sin ^2}x - \sin x - 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = \frac{{1 + \sqrt {17} }}{4}\,\,\,\,\left( {ktm} \right)\\ \sin x = \frac{{1 - \sqrt {17} }}{4} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \arcsin \frac{{1 - \sqrt {17} }}{4} + k2\pi \,\\ x = \pi - \arcsin \frac{{1 - \sqrt {17} }}{4} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\]