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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
pt <=> 1+cos2x + cos3x + cosx = 0
<=> 2cos²x + 2cos2x.cosx = 0
<=> 2cosx.(cos2x + cosx) = 0
<=> 4cosx.cos(3x/2).cos(x/2) = 0 <=>
[cosx = 0
[cos(3x/2) = 0 (tập nghiệm cos3x/2 = 0 chứa tập nghiệm cosx/2 = 0)
<=>
[x = pi/2 + kpi
[3x/2 = pi/2 + kpi
<=>
[x = pi/2 + kpi
[x = pi/3 + 2kpi/3 (k thuộc Z)
sin^2 x + sin^2 2x + sin^2 3x + sin^2 4x =
[1-cos(2x)]/2+ [1-cos(4x)]/2+[1-cos(6x)]/2+[1-cos(8x)]/... =
2- [ cos(2x)+cos(4x)+cos(6x)+cos(8x)]/2 =
2- 1/2· [ cos(2x)+cos(8x)]+cos(4x)+cos(6x)]=
2- 1/2· [ 2·cos(-3x)·cos(5x) + 2· cos(-x)·cos(5x)]=
2- cos(5x)· [cos(3x)+cosx] =
2- cos(5x)· 2·cos(2x)·cosx =
2- 2·cosx·cos(2x)·cos(5x)= 2 <-->
*cosx=0 --> x= pi/2+ k·pi with k thuộc Z or
*cos(2x)=0 --> x= pi/4 + k·pi/2 with k thuộc Z or
* cos(5x)=0 --> x= pi/10+ k·pi/5 with k thuộc Z
e.
\(3\left(1-sin^2x\right)-5sinx-1=0\)
\(\Leftrightarrow-3sin^2x-5sinx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{3}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
f.
\(2\left(2cos^2x-1\right)-cosx+7=0\)
\(\Leftrightarrow4cos^2x-cosx+5=0\)
Phương trình vô nghiệm
g.
\(\Leftrightarrow\sqrt{2}sin\left(4x+\frac{\pi}{4}\right)=2\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=\sqrt{2}>1\)
Phương trình vô nghiệm
h.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
1/ \(pt\Leftrightarrow\left(3cos^2x-sin^2x\right)\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow\left(\dfrac{3}{2}\left(1+cos2x\right)-\dfrac{1}{2}\left(1-cos2x\right)\right)\left(\dfrac{1}{2}\left(1+cos2x\right)-\dfrac{1}{2}\left(1-cos2x\right)\right)=0\)
\(\Leftrightarrow\left(2cos2x+1\right)cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
2/ \(pt\Leftrightarrow\left(sinx-1\right)\left(sin^2x+sinx+6\right)=0\)
\(\Leftrightarrow sinx=1\)
3/ \(pt\Leftrightarrow\dfrac{1-cos2x}{2}-4sin2x+\dfrac{7}{2}\left(1+cos2x\right)=0\)
\(\Leftrightarrow3cos2x-4sin2x=-4\)
\(\Leftrightarrow5\left(\dfrac{3}{5}cos2x-\dfrac{4}{5}sin2x\right)=-4\)
\(\Leftrightarrow cos\left(2x+arccos\dfrac{3}{5}\right)=-\dfrac{4}{5}\)
4,5 giải tương tự câu 3
\(\Leftrightarrow2cosx-sinx-4sin^2x.cosx+2sin^3x=sin^3x+cos^3x\)
\(\Leftrightarrow sin^3x-cos^3x-4sin^2x.cosx+2cosx-sinx=0\)
- Với \(\left\{{}\begin{matrix}cosx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\) là nghiệm của pt
- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)
\(tan^3x-1-4tan^2x+2\left(1+tan^2x\right)-tanx\left(1+tan^2x\right)=0\)
\(\Leftrightarrow-2tan^2x-tanx+3=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{3}{2}\right)+k\pi\end{matrix}\right.\)
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.