Toán Gpt cos2x/cosx=tanx .vieet ra giấy giúp e vs ạ 11/09/2021 By Eden Gpt cos2x/cosx=tanx .vieet ra giấy giúp e vs ạ
Giải thích các bước giải: $\begin{array}{l} DKXD:\cos x \ne 0 \Rightarrow x \ne \frac{\pi }{2} + k\pi \left( {k \in Z} \right)\\ \frac{{\cos 2x}}{{\cos x}} = \tan x\\ \Rightarrow \frac{{\cos 2x}}{{\cos x}} – \frac{{\sin x}}{{\cos x}} = 0\\ \Rightarrow \cos 2x – \sin x = 0\\ \Rightarrow 1 – 2{\sin ^2}x – \sin x = 0\\ \Rightarrow \left( { – 2\sin x + 1} \right)\left( {\sin x + 1} \right) = 0\\ \Rightarrow \left[ \begin{array}{l} \sin x = \frac{1}{2}\\ \sin x = – 1 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \frac{\pi }{6} + k2\pi \left( {tm} \right)\\ x = \frac{{5\pi }}{6} + k2\pi \left( {tm} \right)\\ x = \frac{{ – \pi }}{2} + k2\pi \,\left( {ktm} \right) \end{array} \right.\\ Vay\,x = \frac{\pi }{6} + k2\pi \,hoac\,x = \frac{{5\pi }}{6} + k2\pi \end{array}$ Trả lời
Giải thích các bước giải:
$\begin{array}{l}
DKXD:\cos x \ne 0 \Rightarrow x \ne \frac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
\frac{{\cos 2x}}{{\cos x}} = \tan x\\
\Rightarrow \frac{{\cos 2x}}{{\cos x}} – \frac{{\sin x}}{{\cos x}} = 0\\
\Rightarrow \cos 2x – \sin x = 0\\
\Rightarrow 1 – 2{\sin ^2}x – \sin x = 0\\
\Rightarrow \left( { – 2\sin x + 1} \right)\left( {\sin x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin x = \frac{1}{2}\\
\sin x = – 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \left( {tm} \right)\\
x = \frac{{5\pi }}{6} + k2\pi \left( {tm} \right)\\
x = \frac{{ – \pi }}{2} + k2\pi \,\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = \frac{\pi }{6} + k2\pi \,hoac\,x = \frac{{5\pi }}{6} + k2\pi
\end{array}$
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