cho 3sin^4x – cos^4x = 1/2. tính A = 2sin^4x – cos^4x 02/08/2021 Bởi Madelyn cho 3sin^4x – cos^4x = 1/2. tính A = 2sin^4x – cos^4x
sin2x+cos2x=1⇔cos2x=1−sin2x3sin4x−cos4x=12⇔3sin4x−(1−sin2x)2=12⇔3sin4x−1+2sin2x−sin4x−12=0⇔2sin4x+2sin2x−32=0⇔[sin2x=12sin2x=−32⇒sin2x=12⇒cos2x=12A=2sin4x−cos4x=2.(12)2−(12)2=14 Bình luận
Đáp án: \[A = \frac{1}{4}\] Giải thích các bước giải: Ta có: \[\begin{array}{l}{\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\cos ^2}x = 1 – {\sin ^2}x\\3{\sin ^4}x – {\cos ^4}x = \frac{1}{2}\\ \Leftrightarrow 3{\sin ^4}x – {\left( {1 – {{\sin }^2}x} \right)^2} = \frac{1}{2}\\ \Leftrightarrow 3{\sin ^4}x – 1 + 2{\sin ^2}x – {\sin ^4}x – \frac{1}{2} = 0\\ \Leftrightarrow 2{\sin ^4}x + 2{\sin ^2}x – \frac{3}{2} = 0\\ \Leftrightarrow \left[ \begin{array}{l}{\sin ^2}x = \frac{1}{2}\\{\sin ^2}x = – \frac{3}{2}\end{array} \right. \Rightarrow {\sin ^2}x = \frac{1}{2} \Rightarrow {\cos ^2}x = \frac{1}{2}\\A = 2{\sin ^4}x – {\cos ^4}x = 2.{\left( {\frac{1}{2}} \right)^2} – {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\end{array}\] Bình luận
sin2x+cos2x=1⇔cos2x=1−sin2x3sin4x−cos4x=12⇔3sin4x−(1−sin2x)2=12⇔3sin4x−1+2sin2x−sin4x−12=0⇔2sin4x+2sin2x−32=0⇔[sin2x=12sin2x=−32⇒sin2x=12⇒cos2x=12A=2sin4x−cos4x=2.(12)2−(12)2=14
Đáp án:
\[A = \frac{1}{4}\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
{\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\cos ^2}x = 1 – {\sin ^2}x\\
3{\sin ^4}x – {\cos ^4}x = \frac{1}{2}\\
\Leftrightarrow 3{\sin ^4}x – {\left( {1 – {{\sin }^2}x} \right)^2} = \frac{1}{2}\\
\Leftrightarrow 3{\sin ^4}x – 1 + 2{\sin ^2}x – {\sin ^4}x – \frac{1}{2} = 0\\
\Leftrightarrow 2{\sin ^4}x + 2{\sin ^2}x – \frac{3}{2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = \frac{1}{2}\\
{\sin ^2}x = – \frac{3}{2}
\end{array} \right. \Rightarrow {\sin ^2}x = \frac{1}{2} \Rightarrow {\cos ^2}x = \frac{1}{2}\\
A = 2{\sin ^4}x – {\cos ^4}x = 2.{\left( {\frac{1}{2}} \right)^2} – {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}
\end{array}\]