Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
\(\Leftrightarrow4sinx.cosx+3\left(sinx-cosx\right)=0\)
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(2\left(1-t^2\right)+3t=0\)
\(\Leftrightarrow-2t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=2\left(l\right)\\t=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow sinx-cosx=-\frac{1}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{1}{2\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
2.
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-4t=4\)
\(\Leftrightarrow t^2+4t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sinx-cosx=-1\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{3\pi}{2}+k2\pi\end{matrix}\right.\)
1.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3=1+\frac{1-t^2}{2}\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
1.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx+\frac{\sqrt{2}}{2}=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)+\frac{\sqrt{2}}{2}=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{17\pi}{12}+k2\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\frac{3}{\sqrt{13}}sin2x+\frac{2}{\sqrt{13}}cos2x=\frac{3}{\sqrt{13}}\)
Đặt \(\frac{3}{\sqrt{13}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sin2x.cosa+cos2x.sina=cosa\)
\(\Leftrightarrow sin\left(2x+a\right)=sin\left(\frac{\pi}{2}-a\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+a=\frac{\pi}{2}-a+k2\pi\\2x+a=\frac{\pi}{2}+a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}-a+k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}\)
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{3}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{12}+k2\pi\\x=\frac{13\pi}{12}+k2\pi\end{matrix}\right.\)
4.
Câu này giống hệt câu a
a/ \(4cos^3x-3cosx-4\left(2cos^2x-1\right)+3cosx-4=0\)
\(\Leftrightarrow4cos^3x-8cos^2x=0\)
\(\Leftrightarrow4cos^2x\left(cosx-2\right)=0\)
\(\Leftrightarrow cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)
\(0< \frac{\pi}{2}+k\pi< 14\Rightarrow-\frac{1}{2}< k< \frac{14-\frac{\pi}{2}}{\pi}\Rightarrow k=\left\{0;1;2;3\right\}\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\frac{3\pi}{2};\frac{5\pi}{2};\frac{7\pi}{2}\right\}\)
b/ Bạn coi lại đề, cái ngoặc thứ 2 thiếu \(\left(2cos\left(???\right)+cosx\right)\)
c/ Bạn coi lại đề, có 2 số hạng \(cos2x\) xuất hiện ở vế trái, cấp 3 chắc ko ai cho kiểu vậy đâu, nếu đúng thế thì người ta cộng luôn thành \(2cos2x\) cho rồi
1.
\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}+\sqrt{3}cos\frac{x}{2}=0\)
\(\Leftrightarrow cos\frac{x}{2}\left(2sin\frac{x}{2}+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{x}{2}=-\frac{\pi}{3}+k2\pi\\\frac{x}{2}=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)
2.
\(\Leftrightarrow cosx=2cos^2\left(\frac{x}{2}-\frac{\pi}{6}\right)-1\)
\(\Leftrightarrow cosx=cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=x-\frac{\pi}{3}+k2\pi\left(vn\right)\\x=\frac{\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{6}+k\pi\)
3.
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow x-\frac{\pi}{3}=k\pi\)
\(\Leftrightarrow...\)
4.
\(1+\frac{1}{2}sin6x=sin^2x+cos^2x+2sinx.cosx\)
\(\Leftrightarrow\frac{1}{2}sin6x=sin2x\)
\(\Leftrightarrow sin6x-2sin2x=0\)
\(\Leftrightarrow3sin2x-4sin^32x-2sin2x=0\)
\(\Leftrightarrow sin2x-4sin^32x=0\)
\(\Leftrightarrow sin2x\left(1-4sin^22x\right)=0\)
\(\Leftrightarrow sin2x\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
4.
\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+2sin^2x+3sinx-2=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\frac{\pi}{4}-x=-\frac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)
3.
\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)
\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
c/
\(\Leftrightarrow1+2cos^2x-1+cosx=0\)
\(\Leftrightarrow2cos^2x-cosx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
d/
Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\ge0\\cosx=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+3b=2\\a^2+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2-3b\\a^2+b^2=1\end{matrix}\right.\)
\(\Rightarrow\left(2-3b\right)^2+b^2-1=0\)
\(\Rightarrow10b^2-12b+3=0\Rightarrow\left[{}\begin{matrix}b=\frac{6+\sqrt{6}}{10}\Rightarrow a=\frac{2-3\sqrt{6}}{10}\left(l\right)\\b=\frac{6-\sqrt{6}}{10}\Rightarrow a=\frac{2+3\sqrt{6}}{10}\end{matrix}\right.\)
\(\Rightarrow cosx=\frac{6-\sqrt{6}}{10}\)
\(\Rightarrow x=\pm arccos\left(\frac{6-\sqrt{6}}{10}\right)+k2\pi\)
b/
\(cos\left(8sinx\right)=1\)
\(\Leftrightarrow8sinx=k2\pi\)
\(\Leftrightarrow sinx=\frac{k\pi}{4}\)
Do \(-1\le sinx\le1\Rightarrow-1\le\frac{k\pi}{4}\le1\)
\(\Rightarrow k=\left\{-1;0;1\right\}\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{\pi}{4}\\sinx=0\\sinx=\frac{\pi}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=\pi\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=k\pi\end{matrix}\right.\)