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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.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
a. ĐKXĐ: ...
\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)
b.
\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)
\(\Leftrightarrow4cos^32x-2cos2x-1=0\)
Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề
c. ĐKXĐ: ...
\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Rightarrow x=...\\\frac{cosx+sinx}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cosx+sinx=sinx.cosx\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)
c.
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)
d.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
a.
\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)
\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
a, ta có 2x + π/3 = 3π/4 +k2π hoặc 2x + π/3 = -3π/4 + k2π
=> x= 5π/24 + kπ hoặc x= -13π/24 +kπ
b, đề sai phải ko
c, cos22x - sin22x - 2sinx -1=0
<=> -2sin22x -2sin2x =0
<=> sin2x=0 hoặc sin2x=-1
<=> x=kπ hoặc x= π/2 + kπ ; x=-π/4 +kπ hoặc x=5π/8 + kπ
d, cos5xcosπ/4 - sin5xsinπ/4 = -1/2
cos( 5x + π/4 ) = -1/2
<=> x=π/12 +k2π/5 hoặc x= -11π/60 + k2π/5
f,4x+π/3=3π/10 -x +k2π hoặc 4x+π/3 = x - 3π/10 +k2π
<=> x =-π/150 + k2π/5 hoặc x = π/90 +k2π/3
a.
\(\Leftrightarrow\left[{}\begin{matrix}3x=90^0-x+k360^0\\3x=90^0+x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{45^0}{2}+k90^0\\x=45^0+k180^0\end{matrix}\right.\)
b.
\(\Leftrightarrow cos\left(3x+45^0\right)=cos\left(x-180^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+45^0=x-180^0+k360^0\\3x+45^0=180^0-x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{225^0}{2}+k180^0\\x=\frac{135^0}{4}+k90^0\end{matrix}\right.\)
c.
\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=-x+k2\pi\\2x+\frac{\pi}{3}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
d.
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=cos2x\)
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2\pi}{3}=\frac{\pi}{2}-x+k2\pi\\x-\frac{2\pi}{3}=2x+\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{12}+k\pi\\x=-\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
e.
\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow2x-\frac{\pi}{4}=\frac{\pi}{6}-2x+k2\pi\)
\(\Leftrightarrow x=\frac{5\pi}{48}+\frac{k\pi}{2}\)
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{1}{cos^2x}\left(9-13cosx\right)+4=0\)
\(\Leftrightarrow\frac{9}{cos^2x}-\frac{13}{cosx}+4=0\)
Đặt \(\frac{1}{cosx}=t\)
\(\Rightarrow9t^2-13t+4=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{4}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=1\\\frac{1}{cosx}=\frac{4}{9}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
d/
\(\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0\)
\(\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^22x+sin2x+1=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)
\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)
\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)
\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)
\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)
a)
\(4\sin (3x+\frac{\pi}{3})-2=0\Leftrightarrow \sin (3x+\frac{\pi}{3})=\frac{1}{2}=\sin (\frac{\pi}{6})\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{\pi}{3}=\frac{\pi}{6}+2k\pi \\ 3x+\frac{\pi}{3}=\pi-\frac{\pi}{6}+2k\pi\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=\frac{-\pi}{18}+\frac{2\pi}{3}\\ x=\frac{\pi}{6}+\frac{2\pi}{3}\end{matrix}\right.\) (k nguyên)
c)
\(\sin (x+\frac{x}{4})-1=0\Leftrightarrow \sin (\frac{5}{4}x)=1=\sin (\frac{\pi}{2})\)
\(\Rightarrow \frac{5}{4}x=\frac{\pi}{2}+2k\pi\Rightarrow x=\frac{2}{5}\pi+\frac{8}{5}k\pi \) (k nguyên)
d)
\(2\sin (2x+70^0)+1=0\Leftrightarrow \sin (2x+\frac{7}{18}\pi)=-\frac{1}{2}=\sin (\frac{-\pi}{6})\)
\(\Rightarrow \left[\begin{matrix} 2x+\frac{7}{18}\pi=\frac{-\pi}{6}+2k\pi\\ 2x+\frac{7}{18}\pi=\frac{7}{6}\pi+2k\pi\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=\frac{-5\pi}{18}+k\pi\\ x=\frac{7}{18}\pi+k\pi\end{matrix}\right.\)
f)
\(\cos 2x-\cos 4x=0\)
\(\Leftrightarrow \cos 2x=\cos 4x\Rightarrow \left[\begin{matrix} 4x=2x+2k\pi\\ 4x=-2x+2k\pi\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=k\pi\\ x=\frac{k}{3}\pi \end{matrix}\right.\) ( k nguyên)
b,e,g bạn xem lại đề, đơn vị không thống nhất.