Chủ đề này có 1 trả lời

[htatgiang]

$\frac{1}{sinx}+\frac{1}{sin\left ( x-\frac{3\pi }{2} \right )}=4sin\left ( \frac{7\pi }{4}-x \right )$

[sieumau88]

$\dfrac{1}{sinx}+\dfrac{1}{sin\left ( x-\dfrac{3\pi }{2} \right )}=4sin\left ( \dfrac{7\pi }{4}-x \right )$

 

$\dfrac{1}{sinx}+\dfrac{1}{sin\left ( x-\dfrac{3\pi }{2} \right )}=4sin\left ( \dfrac{7\pi }{4}-x \right )$

 

$\Leftrightarrow$ $\dfrac{1}{sinx}+\dfrac{1}{cosx}= 4sin\left ( -\dfrac{\pi }{4}-x \right )$

 

ĐK : $\left\{\begin{matrix}
sinx\neq 0 & \\
cosx \neq 0 &
\end{matrix}\right.$ $\Leftrightarrow$__$sin2x \neq  0$ __$\Leftrightarrow$ _$x \neq  \dfrac{k\pi}{2}$_;_$k \in \mathbb{Z}$

 

pt $\Leftrightarrow$ $\dfrac{cosx+sinx}{sinx.cosx}= -4sin\left ( \dfrac{\pi }{4}+x \right )$

 

$\Leftrightarrow$ $\dfrac{sin\left ( \dfrac{\pi }{4}+x \right )}{\dfrac{sin2x}{2}}= -4sin\left ( \dfrac{\pi }{4}+x \right )$

 

$\Leftrightarrow$ $sin\left ( \dfrac{\pi }{4}+x \right ) = -2sin2x. sin\left ( \dfrac{\pi }{4}+x \right )$

 

$\Leftrightarrow$ $\left[ \begin{array}{l} sin\left ( \dfrac{\pi }{4}+x \right ) = 0 \\ sin2x = \dfrac{-1}{2} \\ \end{array} \right.$

 

$\rightarrow$ ....... v......v.........