Giải phương trình thuần bậc 2 vs sinx và cosx 2sin²x-5sinxcosx-cos²x=-2 16/07/2021 Bởi Genesis Giải phương trình thuần bậc 2 vs sinx và cosx 2sin²x-5sinxcosx-cos²x=-2
Đáp án: $\begin{array}{l}2{\sin ^2}x – 5\sin x.\cos x – {\cos ^2}x = – 2\\ + Khi:\cos x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi \\ \Rightarrow 2{\sin ^2}x = – 2\\ \Rightarrow {\sin ^2}x = – 1\left( {ktm} \right)\\ + Khi:\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi \\ \Rightarrow \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}x}} – \dfrac{{5\sin x.\cos x}}{{{{\cos }^2}x}} – 1 = \dfrac{{ – 2}}{{{{\cos }^2}x}}\\ \Rightarrow 2{\tan ^2}x – 5\tan x – 1 = – 2.\left( {{{\tan }^2}x + 1} \right)\\ \Rightarrow 4{\tan ^2}x – 5\tan x + 1 = 0\\ \Rightarrow \left( {4\tan x – 1} \right)\left( {\tan x – 1} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}\tan x = 1\\\tan x = \dfrac{1}{4}\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\pi \left( {tmdk} \right)\\x = \arctan \dfrac{1}{4} + k\pi \end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
2{\sin ^2}x – 5\sin x.\cos x – {\cos ^2}x = – 2\\
+ Khi:\cos x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi \\
\Rightarrow 2{\sin ^2}x = – 2\\
\Rightarrow {\sin ^2}x = – 1\left( {ktm} \right)\\
+ Khi:\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi \\
\Rightarrow \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}x}} – \dfrac{{5\sin x.\cos x}}{{{{\cos }^2}x}} – 1 = \dfrac{{ – 2}}{{{{\cos }^2}x}}\\
\Rightarrow 2{\tan ^2}x – 5\tan x – 1 = – 2.\left( {{{\tan }^2}x + 1} \right)\\
\Rightarrow 4{\tan ^2}x – 5\tan x + 1 = 0\\
\Rightarrow \left( {4\tan x – 1} \right)\left( {\tan x – 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \left( {tmdk} \right)\\
x = \arctan \dfrac{1}{4} + k\pi
\end{array} \right.
\end{array}$