Câu hỏi:
Giúp em bài tập về nhà Toán lớp 11 câu hỏi như sau: GPT
tanx – tan2x = sinx
Trả lời 1:
Gia Sư Hoàng Khang gữi câu trả lời dành cho bạn:
Giải đáp:
\[x = k\pi \,\,\,\,\left( {k \in \,Z} \right)\]
Lời giải và giải thích chi tiết:
ĐKXĐ:
\(\left\{ \begin{array}{l}
\cos x \ne 0\\
\cos 2x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{2} + k\pi \\
2x \ne \dfrac{\pi }{2} + k\pi
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{2} + k\pi \\
x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\)
Ta có:
\(\begin{array}{l}
\tan x – \tan 2x = \sin x\\
\Leftrightarrow \dfrac{{\sin x}}{{\cos x}} – \dfrac{{\sin 2x}}{{\cos 2x}} = \sin x\\
\Leftrightarrow \dfrac{{\sin x}}{{\cos x}} – \dfrac{{2\sin x.\cos x}}{{2{{\cos }^2}x – 1}} = \sin x\\
\Leftrightarrow \sin x.\left[ {\dfrac{1}{{\cos x}} – \dfrac{{2\cos x}}{{2{{\cos }^2}x – 1}} – 1} \right] = 0\\
\Leftrightarrow \sin x.\dfrac{{\left( {2{{\cos }^2}x – 1} \right) – 2\cos x.\cos x – \cos x.\left( {2{{\cos }^2}x – 1} \right)}}{{\cos x.\left( {2{{\cos }^2}x – 1} \right)}} = 0\\
\Leftrightarrow \sin x.\dfrac{{2{{\cos }^2}x – 1 – 2{{\cos }^2}x – 2{{\cos }^3}x + \cos x}}{{\cos x\left( {2{{\cos }^2}x – 1} \right)}} = 0\\
\Leftrightarrow \sin x.\dfrac{{ – 2{{\cos }^3}x + \cos x – 1}}{{\cos x\left( {2{{\cos }^2}x – 1} \right)}} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
– 2{\cos ^3}x + \cos x – 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\left( { – 2{{\cos }^3}x – 2{{\cos }^2}x} \right) + \left( {2{{\cos }^2}x + 2\cos x} \right) – \left( {\cos x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
– 2{\cos ^2}x\left( {\cos x + 1} \right) + 2\cos x\left( {\cos x + 1} \right) – \left( {\cos x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\left( {\cos x + 1} \right)\left( { – 2{{\cos }^2}x + 2\cos x – 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = – 1\\
– 2{\cos ^2}x + 2\cos x – 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = – 1
\end{array} \right.\\
\Leftrightarrow \sin x = 0\\
\Leftrightarrow x = k\pi
\end{array}\)
Vậy \(x = k\pi \,\,\,\,\left( {k \in \,Z} \right)\)
