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6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
d/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cos^2x+sin^2x+sinx.cosx\right)}{2cosx+3sinx}=cos^2x-sin^2x\)
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(1+sinx.cosx\right)}{2cosx+3sinx}=\left(cosx-sinx\right)\left(cosx+sinx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\\\frac{1+sinx.cosx}{2cosx+3sinx}=sinx+cosx\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1+sinx.cosx=\left(sinx+cosx\right)\left(2cosx+3sinx\right)\)
\(\Leftrightarrow1+sinx.cosx=2sin^2x+3cos^2x+5sinx.cosx\)
\(\Leftrightarrow2sin^2x+3cos^2x+4sinx.cosx-1=0\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(2tan^2x+3+4tanx-1-tan^2x=0\)
\(\Leftrightarrow tan^2x+4tanx+2=0\)
\(\Leftrightarrow tanx=-2\pm\sqrt{2}\)
\(\Rightarrow x=arctan\left(-2\pm\sqrt{2}\right)+k\pi\)
c/
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx\right)=4\left(sinx-cosx\right)\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\sinx+4cosx-4=0\left(2\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
Xét (2) \(\Leftrightarrow\frac{1}{\sqrt{17}}sinx+\frac{4}{\sqrt{17}}cosx=\frac{4}{\sqrt{17}}\)
Đặt \(\frac{4}{\sqrt{17}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow cosx.cosa+sinx.sina=cosa\)
\(\Leftrightarrow cos\left(x-a\right)=cosa\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=-a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=k2\pi\end{matrix}\right.\)
5.
\(\Leftrightarrow sin\left(2cosx\right)=1\)
\(\Leftrightarrow2cosx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow cosx=\frac{\pi}{4}+k\pi\)
Do \(-1\le cosx\le1\Rightarrow-1\le\frac{\pi}{4}+k\pi\le1\)
Mà \(k\in Z\Rightarrow k=0\)
\(\Rightarrow cosx=\frac{\pi}{4}\)
\(\Leftrightarrow x=\pm arccos\left(\frac{\pi}{4}\right)+k2\pi\)
3.
\(\Leftrightarrow sin2x+1=2\left(\frac{1-cos2x}{2}\right)\)
\(\Leftrightarrow sin2x+cos2x=0\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow2x+\frac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)
4. ĐKXĐ; ...
\(\Leftrightarrow\frac{sinx.cos2x}{cosx.sin2x}+1=0\)
\(\Leftrightarrow sinx.cos2x+cosx.sin2x=0\)
\(\Leftrightarrow sin3x=0\)
\(\Leftrightarrow3sinx-4sin^3x=0\)
\(\Leftrightarrow3-4sin^2x=0\)
\(\Leftrightarrow3-2\left(1-cos2x\right)=0\)
\(\Leftrightarrow cos2x=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
Câu 1:
\(cos^2\) gì nhỉ?
Câu 2: đề không hợp lý \(\sqrt{3}sin9x\) là \(\sqrt{3}cos9x\) có lý hơn
\(\Leftrightarrow3sin3x-4sin^33x+\sqrt{3}sin9x=1\)
\(\Leftrightarrow sin9x+\sqrt{3}sin9x=1\)
\(\Leftrightarrow sin9x=\frac{1}{\sqrt{3}+1}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{9}arcsin\left(\frac{1}{\sqrt{3}+1}\right)+\frac{k2\pi}{9}\\x=\frac{\pi}{9}-\frac{1}{9}arcsin\left(\frac{1}{\sqrt{3}+1}\right)+\frac{k2\pi}{9}\end{matrix}\right.\)
Nghiệm nhìn rất ngớ ngẩn nếu đề đúng
3.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\2x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
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
⇔ 
\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)
\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)
\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)
\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)
\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)
2/
\(\Leftrightarrow3sinx-4sin^3x-\sqrt{3}cosx=2sinx\)
\(\Leftrightarrow4sin^3x-sinx+\sqrt{3}cosx=0\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow4tan^3x-tanx\left(1+tan^2x\right)+\sqrt{3}\left(1+tan^2x\right)=0\)
\(\Leftrightarrow3tan^3x+\sqrt{3}tan^2x-tanx+\sqrt{3}=0\)
Bạn xem lại đề, pt bậc 3 này ko giải được (nghiệm rất xấu)
1.
\(\Leftrightarrow\sqrt{3}cos^2x-\sqrt{3}+cos^2x+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)
\(\Leftrightarrow-\sqrt{3}sin^2x+cosx+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+\sqrt{3}sinx\right)-\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+\sqrt{3}sinx-1\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx-\frac{1}{2}\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\left[sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)
mik lm biếng quá mik chỉ nói cách làm thôi nha bạn
1) chia hai vế cho cos^2(x) \(\sqrt{3}tan^2x+\left(1-\sqrt{3}\right)tanx-1+\left(1-\sqrt{3}\right)\left(1+tan^2x\right)=0\)
đặt t = tanx rr giải thôi =D ( máy 570 thì mode5 3 còn máy 580 thì mode 9 2 2) :)))
2) cx làm cách tương tự chia 2 vế cho cos^2x
3) giữ vế trái bung vế phải ra
\(sin2x-2sin^2x=2-4sin^22x\)
đặt t = sin2x (-1=<t=<1)
4) đẩy sinx cosx qua trái hết
\(sinx\left(sin^2-1\right)-cosx\left(cos^2x+1\right)=0\)
\(sinx\left(-cos^2x\right)-cos\left(cos^2x+1\right)=0\)
\(-cos\left(sinxcosx+cos^2x+1\right)=0\)
cái vế đầu cosx=0 bn bik giả rr mà dễ ẹc à còn vế sau thì chia cho cos^2(x) như mấy bài trên rr sau đó đặt t = tanx rr bấm máy là ra thui :))
5)bung cái hằng đẳng thức ra sau đó đặt t=sinx+cosx (t thuộc [-căn(2) ; căn(2)]
khi đó ta có sinxcosx=1/2 sin2x= 1/2t^2 - 1/2
làm đi là ra à
a/
\(sin^2x-sinx=2\left(1-sin^2x\right)\)
\(\Leftrightarrow3sin^2x-sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=arcsin\left(\frac{2}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{2}{3}\right)+k2\pi\end{matrix}\right.\)
2.
\(2sin^2x+\left(1-\sqrt{3}\right)sinx-\frac{\sqrt{3}}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{4}=\frac{\pi}{8}+k2\pi\\3x+\frac{\pi}{4}=-\frac{\pi}{8}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+\frac{k2\pi}{3}\\x=-\frac{\pi}{8}+\frac{k2\pi}{3}\end{matrix}\right.\)
\(cos^2x-sin^2x-2\sqrt{3}sinxcosx=1\)
\(\Leftrightarrow cos2x-\sqrt{3}sin2x=1\)
\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)
\(\Leftrightarrow cos\left(\frac{\pi}{3}\right)cos2x-sin\left(\frac{\pi}{3}\right)sin2x=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)
\(\Leftrightarrow2x+\frac{\pi}{3}=\pm\frac{\pi}{3}+k2\pi\left(k\inℤ\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=k\pi\\x=\frac{-\pi}{3}+k\pi\end{cases}}\left(k\inℤ\right)\)
\(cos^2x-sin^2x-2\sqrt{3}sinx.cosx=1\)
\(\Leftrightarrow cos2x-\sqrt{3}sin2x=1\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\Leftrightarrow2x+\frac{\pi}{3}=\pm\frac{\pi}{3}+k2\pi\)
\(\Leftrightarrow\orbr{\begin{cases}x=k\pi\\x=-\frac{\pi}{3}+k\pi\end{cases}}\)