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2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a) Đk: sinx \(\ne\)0<=>x\(\ne\)k\(\Pi\)
pt<=>\(\sqrt{3}\)(1-cos2x)-cosx=0
<=>\(\sqrt{3}\)[1-(2cos2x-1)]-cosx=0
<=>2\(\sqrt{3}\)-2\(\sqrt{3}\)cos2x-cosx=0
<=>\(\left\{{}\begin{matrix}cosx=\dfrac{\sqrt{3}}{2}\\cosx=-\dfrac{2\sqrt{3}}{3}< -1\left(loai\right)\end{matrix}\right.\)
tới đây bạn tự giải cho quen, chứ chép thì thành ra không hiểu gì thì khổ
b)pt<=>2sin2x+2sin2x=1
<=>2sin2x+2sin2x=sin2x+cos2x
<=>4sinx.cosx+sin2x-cos2x=0
Tới đây là dạng của pt đẳng cấp bậc 2, ta thấy cosx=0 không phải là nghiệm của pt nên ta chia cả hai vế của pt cho cos2x:
pt trở thành:
4tanx+tan2x-1=0
<=>\(\left[{}\begin{matrix}tanx=-2+\sqrt{2}\\tanx=-2-\sqrt{5}\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=arctan\left(-2+\sqrt{5}\right)+k\Pi\\x=arctan\left(-2-\sqrt{5}\right)+k\Pi\end{matrix}\right.\)(k thuộc Z)
Chú ý: arctan tương ứng ''SHIFT tan'' (khi thử nghiệm trong máy tính)
c)Đk: cosx\(\ne\)0<=>x\(\ne\)\(\dfrac{\Pi}{2}\)+kpi
pt<=>cos2x+\(\sqrt{3}\)sin2x=1
<=>1-sin2x+\(\sqrt{3}\)sin2x-1=0
<=>(\(\sqrt{3}\)-1)sin2x=0
<=>sinx=0<=>x=k\(\Pi\)(k thuộc Z)
d)
pt<=>\(\sqrt{3}\)sin7x-cos7x=\(\sqrt{2}\)
Khúc này bạn coi SGK trang 35 người ta giả thích rõ ràng rồi
pt<=>\(\dfrac{\sqrt{3}}{2}\)sin7x-\(\dfrac{1}{2}\)cos7x=\(\dfrac{\sqrt{2}}{2}\)
<=>sin(7x-\(\dfrac{\Pi}{3}\))=\(\dfrac{\sqrt{2}}{2}\)
<=>sin(7x-\(\dfrac{\Pi}{3}\))=sin\(\dfrac{\Pi}{4}\)
Tới đây bạn tự giải nhé, giải ra nghiệm rồi kiểm tra xem nghiệm nào thuộc khoảng ( đề cho) rồi kết luận
Câu d) mình nhầm nhé
<=>sin(7x-\(\dfrac{\Pi}{6}\))=\(\dfrac{\sqrt{2}}{2}\) mới đúng sorry
c/
Đặt \(3cosx-4sinx-6=t\)
Pt trở thành:
\(t^2+2=-3t\Leftrightarrow t^2+3t+2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3cosx-4sinx-6=-1\\3cosx-4sinx-6=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3cosx-4sinx=5\\3cosx-4sinx=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx.\frac{3}{5}-sinx.\frac{4}{5}=1\\cosx.\frac{3}{5}-sinx.\frac{4}{5}=\frac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+a\right)=1\\cosx\left(x+a\right)=\frac{4}{5}\end{matrix}\right.\) (với góc \(a\in\left[0;\pi\right]\) sao cho \(cosa=\frac{3}{5}\))
\(\Leftrightarrow\left[{}\begin{matrix}x+a=k2\pi\\x+a=\pm\left(\frac{\pi}{2}-a\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-a+k2\pi\\x=-a\pm\left(\frac{\pi}{2}-a\right)+k2\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow cosx.\frac{1}{2}-\frac{\sqrt{3}}{2}sinx=cos\left(\frac{\pi}{3}-x\right)\)
\(\Leftrightarrow cosx.cos\left(\frac{\pi}{3}\right)-sinx.sin\left(\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}-x\right)\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{3}-x+k2\pi\\x+\frac{\pi}{3}=-\frac{\pi}{3}+x+k2\pi\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=k\pi\)
b/
\(\Leftrightarrow\sqrt{2}sin\left(5x+\frac{\pi}{4}\right)=\sqrt{2}cos13x\)
\(\Leftrightarrow cos\left(\frac{\pi}{4}-5x\right)=cos13x\)
\(\Leftrightarrow\left[{}\begin{matrix}13x=\frac{\pi}{4}-5x+k2\pi\\13x=-\frac{\pi}{4}+5x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{72}+\frac{k\pi}{9}\\x=-\frac{\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)
1.
\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
2.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
1.
\(8sinx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\)
\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}}{2cosx}+\dfrac{1}{2sinx}\)
\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}sinx+cosx}{sin2x}\)
\(\Leftrightarrow4sinx.sin2x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow2cosx-2cos3x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=2cos3x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos3x\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=\pm3x+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}-k\pi\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)
2.
ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(sinx+\sqrt{3}cosx=\dfrac{1}{cosx}\)
\(\Leftrightarrow2sinx.cosx+2\sqrt{3}cos^2x-\sqrt{3}=2-\sqrt{3}\)
\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=1-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{2-\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\\2x+\dfrac{\pi}{3}=\pi-arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\\x=\dfrac{\pi}{3}-\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\end{matrix}\right.\)
3.3 d)
\(\sin8x-\cos6x=\sqrt{3}\left(\sin6x+\cos8x\right)\\ \Leftrightarrow\sin8x-\sqrt{3}\cos8x=\sqrt{3}\sin6x+\cos6x\\ \Leftrightarrow\sin\left(8x-\dfrac{\pi}{3}\right)=\sin\left(6x+\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\pi-\left(6x+\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{7}\end{matrix}\right.\)
3.4 a)
\(2sin\left(x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(-x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \)
Chia hai vế cho \(\sqrt{2^2+4^2}=2\sqrt{5}\)
Ta được:
\(\dfrac{1}{\sqrt{5}}cos\left(x-\dfrac{\pi}{4}\right)+\dfrac{2}{\sqrt{5}}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3}{4}\\ \)
Gọi \(\alpha\) là góc có \(cos\alpha=\dfrac{1}{\sqrt{5}}\)và \(sin\alpha=\dfrac{2}{\sqrt{5}}\)
Phương trình tương đương:
\(cos\left(x-\dfrac{\pi}{4}-\alpha\right)=\dfrac{3}{4}\\ \Leftrightarrow x=\pm arscos\left(\dfrac{3}{4}\right)+\dfrac{\pi}{4}+\alpha+k2\pi\)
