4sin ²2x+8cos ²x -5 =0 dòng số 2 sao lại phải chia cho 2 v mn, e xin luôn công thức vs ạ, 26/07/2021 Bởi Genesis 4sin ²2x+8cos ²x -5 =0 dòng số 2 sao lại phải chia cho 2 v mn, e xin luôn công thức vs ạ,
Đáp án: \[\left[ \begin{array}{l}x = \pm \dfrac{\pi }{3} + k2\pi \\x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\] Giải thích các bước giải: Ta có: \(\begin{array}{l}4{\sin ^2}2x + 8{\cos ^2}x – 5 = 0\\ \Leftrightarrow 4.{\left( {2\sin x.\cos x} \right)^2} + 8.{\cos ^2}x – 5 = 0\\ \Leftrightarrow 16.{\sin ^2}x.{\cos ^2}x + 8{\cos ^2}x – 5 = 0\\ \Leftrightarrow 16.\left( {1 – {{\cos }^2}x} \right).{\cos ^2}x + 8{\cos ^2}x – 5 = 0\\ \Leftrightarrow 16{\cos ^2}x – 16{\cos ^4}x + 8{\cos ^2}x – 5 = 0\\ \Leftrightarrow – 16{\cos ^4}x + 24{\cos ^2}x – 5 = 0\\ \Leftrightarrow 16{\cos ^4}x – 24{\cos ^2}x + 5 = 0\\ \Leftrightarrow \left( {4{{\cos }^2}x – 1} \right)\left( {4{{\cos }^2}x – 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}{\cos ^2}x = \dfrac{1}{4}\\{\cos ^2}x = \dfrac{5}{4}\end{array} \right.\\ – 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\ \Rightarrow {\cos ^2}x = \dfrac{1}{4} \Leftrightarrow \left[ \begin{array}{l}\cos x = \dfrac{1}{2}\\\cos x = – \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \pm \dfrac{\pi }{3} + k2\pi \\x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\end{array}\) Bình luận
Đáp án:
\[\left[ \begin{array}{l}
x = \pm \dfrac{\pi }{3} + k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4{\sin ^2}2x + 8{\cos ^2}x – 5 = 0\\
\Leftrightarrow 4.{\left( {2\sin x.\cos x} \right)^2} + 8.{\cos ^2}x – 5 = 0\\
\Leftrightarrow 16.{\sin ^2}x.{\cos ^2}x + 8{\cos ^2}x – 5 = 0\\
\Leftrightarrow 16.\left( {1 – {{\cos }^2}x} \right).{\cos ^2}x + 8{\cos ^2}x – 5 = 0\\
\Leftrightarrow 16{\cos ^2}x – 16{\cos ^4}x + 8{\cos ^2}x – 5 = 0\\
\Leftrightarrow – 16{\cos ^4}x + 24{\cos ^2}x – 5 = 0\\
\Leftrightarrow 16{\cos ^4}x – 24{\cos ^2}x + 5 = 0\\
\Leftrightarrow \left( {4{{\cos }^2}x – 1} \right)\left( {4{{\cos }^2}x – 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\cos ^2}x = \dfrac{1}{4}\\
{\cos ^2}x = \dfrac{5}{4}
\end{array} \right.\\
– 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\
\Rightarrow {\cos ^2}x = \dfrac{1}{4} \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x = – \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{3} + k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)