4(sin$x^{2}$ + $\frac{1}{sinx^{2}}$) – 4(sinx-$\frac{1}{sinx}$)=7 giải giúp e với

0 bình luận về “4(sin$x^{2}$ + $\frac{1}{sinx^{2}}$) – 4(sinx-$\frac{1}{sinx}$)=7 giải giúp e với”

1. \[\begin{array}{l}
   4\left( {{{\sin }^2}x + \frac{1}{{{{\sin }^2}x}}} \right) – 4\left( {\sin x – \frac{1}{{\sin x}}} \right) = 7\,\,\,\,\left( {DK:\,\,\,\sin x \ne 0} \right)\\
   \Leftrightarrow 4\left( {{{\sin }^2}x – 2 + \frac{1}{{{{\sin }^2}x}} + 2} \right) – 4\left( {\sin x – \frac{1}{{\sin x}}} \right) = 7\\
   \Leftrightarrow 4{\left( {\sin x – \frac{1}{{\sin x}}} \right)^2} + 8 – 4\left( {\sin x – \frac{1}{{\sin x}}} \right) – 7 = 0\\
   \Leftrightarrow 4{\left( {\sin x – \frac{1}{{\sin x}}} \right)^2} – 4\left( {\sin x – \frac{1}{{\sin x}}} \right) + 1 = 0\\
   \Leftrightarrow {\left[ {2\left( {\sin x – \frac{1}{{\sin x}}} \right) – 1} \right]^2} = 0\\
   \Leftrightarrow 2\left( {\sin x – \frac{1}{{\sin x}}} \right) – 1 = 0\\
   \Leftrightarrow \sin x – \frac{1}{{\sin x}} = \frac{1}{2}\\
   \Leftrightarrow 2{\sin ^2}x – 2 = \sin x\\
   \Leftrightarrow 2{\sin ^2}x – \sin x – 2 = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   \sin x = \frac{{1 + \sqrt {17} }}{4}\,\,\,\,\left( {ktm} \right)\\
   \sin x = \frac{{1 – \sqrt {17} }}{4}
   \end{array} \right.\\
   \Leftrightarrow \left[ \begin{array}{l}
   x = \arcsin \frac{{1 – \sqrt {17} }}{4} + k2\pi \,\\
   x = \pi – \arcsin \frac{{1 – \sqrt {17} }}{4} + k2\pi
   \end{array} \right.\,\,\,\left( {k \in Z} \right).
   \end{array}\]

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