Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu g đề thiếu
Câu 2:
\(sin\left(2x+\frac{\pi}{6}\right)=\frac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=arcsin\left(\frac{2}{5}\right)+k2\pi\\2x+\frac{\pi}{6}=\pi-arcsin\left(\frac{2}{5}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{1}{2}arcsin\left(\frac{2}{5}\right)+k\pi\\x=\frac{5\pi}{12}-\frac{1}{2}arcsin\left(\frac{2}{5}\right)+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\approx-0,056\left(rad\right)\\x\approx1,1\left(rad\right)\end{matrix}\right.\)
Đkxđ: \(x\in R\).
\(cos2x-cos3x+cos4x=0\Leftrightarrow\left(cos2x+cos4x\right)-cos3x=0\)
\(\Leftrightarrow2cos3x.cosx-cos3x=0\)
\(\Leftrightarrow cos3x\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\2cos2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cos3x=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(cos3x=0\Leftrightarrow3x=\dfrac{\pi}{2}+k\pi\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)
\(cos2x=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\dfrac{sinB}{sinC}=2cosA\Leftrightarrow sinB=2cosA.sinC\)
\(\Leftrightarrow sinB=sin\left(A+C\right)+sin\left(C-A\right)\)
\(\Leftrightarrow sinB=sin\left(\pi-\left(A+C\right)\right)+sin\left(C-A\right)\)
\(\Leftrightarrow sinB=sinB+sin\left(C-A\right)\)
\(\Leftrightarrow sin\left(C-A\right)=0\) (1)
Do A, C là số đo các góc trong tam giác nên từ (1) suy ra:
\(C=A\) hay tam giác ABC cân.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
\(\frac{a^2}{1-\frac{sin^2x}{cos^2x}}=\frac{sin^2x+a^2-2}{cos^2x-sin^2x}\Leftrightarrow\frac{a^2.cos^2x}{cos^2x-sin^2x}=\frac{sin^2x+a^2-2}{cos^2x-sin^2x}\)
\(\Leftrightarrow a^2\left(1-sin^2x\right)=sin^2x+a^2-2\)
\(\Leftrightarrow\left(a^2+1\right)sin^2x=2\)
\(\Leftrightarrow sin^2x=\frac{2}{a^2+1}\)
Để pt có nghiệm:
\(\Leftrightarrow\left\{{}\begin{matrix}-1< \frac{2}{a^2+1}< 1\\\frac{2}{a^2+1}\ne\frac{\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2>1\\a^2\ne2\sqrt{2}-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a>1\\a< -1\end{matrix}\right.\\a\ne2\sqrt{2}-1\end{matrix}\right.\)