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c/
\(\Leftrightarrow1-sin^22x+\sqrt{3}sin2x+sin2x=1+\sqrt{3}\)
\(\Leftrightarrow-sin^22x+\left(\sqrt{3}+1\right)sin2x-\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\sqrt{3}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
d/
\(\Leftrightarrow4\left(1-2sin^2x\right)+5sinx=4\left(3sinx-4sin^3x\right)+5\)
\(\Leftrightarrow16sin^3x-8sin^2x-7sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(4sinx+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=arcsin\left(-\frac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{4}\right)+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow3cos^2x+4sin\left(2\pi-\frac{\pi}{2}-x\right)+1=0\)
\(\Leftrightarrow3cos^2x-4sin\left(x+\frac{\pi}{2}\right)+1=0\)
\(\Leftrightarrow3cos^2x-4cosx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm arcos\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
a) Dễ thấy cosx = 0 không thỏa mãn phương trình đã cho nên chiaw phương trình cho cos2x ta được phương trình tương đương 2tan2x + tanx - 3 = 0.
Đặt t = tanx thì phương trình này trở thành
2t2 + t - 3 = 0 ⇔ t ∈ {1 ; 
}.
Vậy 
b) Thay 2 = 2(sin2x + cos2x), phương trình đã cho trở thành
3sin2x - 4sinxcosx + 5cos2x = 2sin2x + 2cos2x
⇔ sin2x - 4sinxcosx + 3cos2x = 0
⇔ tan2x - 4tanx + 3 = 0
⇔ 
⇔ x = 
+ kπ ; x = arctan3 + kπ, k ∈ Z.
c) Thay sin2x = 2sinxcosx ; 
= (sin2x + cos2x) vào phương trình đã cho và rút gọn ta được phương trình tương đương
sin2x + 2sinxcosx - 
cos2x = 0 ⇔ tan2x + 4tanx - 5 = 0 ⇔ 
⇔ x = + kπ ; x = arctan(-5) + kπ, k ∈ Z.
d) 2cos2x - 3√3sin2x - 4sin2x = -4
⇔ 2cos2x - 3√3sin2x + 4 - 4sin2x = 0
⇔ 6cos2x - 6√3sinxcosx = 0 ⇔ cosx(cosx - √3sinx) = 0
⇔ 
c/
\(\Leftrightarrow sin3x-\sqrt{3}cos3x=sinx+\sqrt{3}cosx\)
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=x+\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{3}=\frac{2\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
a/
\(\Leftrightarrow\sqrt{3}cos2x-\left(sin^2x+cos^2x-2sinx.cosx\right)=2\)
\(\Leftrightarrow\sqrt{3}cos2x-1+sin2x=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x=\frac{3}{2}\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=\frac{3}{2}\)
Vế phải lớn hơn 1 nên pt vô nghiệm
b/
\(\Leftrightarrow\frac{5}{2}\left(1+cos2x\right)+2sin2x=4\)
\(\Leftrightarrow4sin2x+5cos2x=3\)
\(\Leftrightarrow\frac{4}{\sqrt{41}}sin2x+\frac{5}{\sqrt{41}}cos2x=\frac{3}{\sqrt{41}}\)
Đặt \(\frac{4}{\sqrt{41}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sin2x.cosa+cos2x.sina=\frac{3}{\sqrt{41}}\)
\(\Leftrightarrow sin\left(2x+a\right)=\frac{3}{\sqrt{41}}=sinb\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+a=b+k2\pi\\2x+a=\pi-b+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{b}{2}-\frac{a}{2}+k\pi\\x=\frac{\pi}{2}-\frac{a}{2}-\frac{b}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow cosx+2sinx+3=2m.cosx-m.sinx+4m\)
\(\Leftrightarrow\left(m+2\right)sinx+\left(1-2m\right)cosx=4m-3\)
Pt đã cho có nghiệm khi và chỉ khi:
\(\left(m+2\right)^2+\left(1-2m\right)^2\ge\left(4m-3\right)^2\)
\(\Leftrightarrow11m^2-24m+4\le0\)
\(\Rightarrow\frac{2}{11}\le m\le2\)
Câu 1 với câu 2 sai đề, sin và cos nằm trong [-1;1], mà căn 2 với căn 3 lớn hơn 1 rồi
3/ \(\sin x=\cos2x=\sin\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}-2x+k2\pi\\x=\pi-\frac{\pi}{2}+2x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\frac{2}{3}\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
4/ \(\Leftrightarrow\cos^2x-2\sin x\cos x=0\)
Xét \(\cos x=0\) là nghiệm của pt \(\Rightarrow x=\frac{\pi}{2}+k\pi\)
\(\cos x\ne0\Rightarrow1-2\tan x=0\Leftrightarrow\tan x=\frac{1}{2}\Rightarrow x=...\)
5/ \(\Leftrightarrow\sin\left(2x+1\right)=-\cos\left(3x-1\right)=\cos\left(\pi-3x+1\right)=\sin\left(\frac{\pi}{2}-\pi+3x-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\frac{\pi}{2}-\pi+3x-1\\2x+1=\pi-\frac{\pi}{2}+\pi-3x+1\end{matrix}\right.\Leftrightarrow....\)
6/ \(\Leftrightarrow\cos\left(\pi\left(x-\frac{1}{3}\right)\right)=\frac{1}{2}\Leftrightarrow\pi\left(x-\frac{1}{3}\right)=\pm\frac{\pi}{3}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{1}{3}+2k\Rightarrow x=\frac{2}{3}+2k\left(1\right)\\x-\frac{1}{3}=-\frac{1}{3}+2k\Rightarrow x=2k\left(2\right)\end{matrix}\right.\)
\(\left(1\right):-\pi< x< \pi\Rightarrow-\pi< \frac{2}{3}+2k< \pi\) (Ủa đề bài sai hay sao ý nhỉ?)
7/ \(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=\frac{\pi}{2}-2x+\frac{\pi}{3}\\5x+\frac{\pi}{3}=\pi-\frac{\pi}{2}+2x-\frac{\pi}{3}\end{matrix}\right.\Leftrightarrow...\)
Thui, để đây bao giờ...hết lười thì làm tiếp :(
7)
\(sin\left(5x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=\frac{\pi}{2}-2x-\frac{\pi}{3}+k2\pi\\5x+\frac{\pi}{3}=\pi-\left(\frac{\pi}{2}-2x-\frac{\pi}{3}\right)+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{42}+k\frac{2\pi}{7}\\x=\frac{\pi}{6}+k\frac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)
Do:\(0< x< \pi\)
\(Với:x=\frac{-\pi}{42}+k\frac{2\pi}{7}\left(k\in Z\right)\Rightarrow khôngtìmđượck\)
\(Với:x=\frac{\pi}{6}+k\frac{2\pi}{3}\left(k\in Z\right)\Leftrightarrow\frac{1}{4}< k< \frac{5}{4}\Rightarrow k=\left\{0;1\right\}\Rightarrow\left[{}\begin{matrix}k=0\Rightarrow x=\frac{\pi}{6}\\k=1\Rightarrow x=\frac{5\pi}{6}\end{matrix}\right.\)
Vậy nghiệm của pt là: \(x=\frac{\pi}{6};x=\frac{5\pi}{6}\)
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:
\(cos\left(x-\dfrac{\pi}{4}-\alpha\right)=\dfrac{3}{4}\\ \Leftrightarrow x=\pm arscos\left(\dfrac{3}{4}\right)+\dfrac{\pi}{4}+\alpha+k2\pi\)
