Giải PT
a) 3cos26x + 8sin 3x cos 3x - 4 =0
b) sinx + 4cos2x + 1 = 0
c) \(\frac{1}{cos^2x}\)+ tanx - 1 = 0
d) sin x + 3sin\(\frac{x}{2}\)= 0
giải pt
a) \(\sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right).tan^2x-cos^2\frac{x}{2}=0\)
b) \(3tan^3x-tanx+\frac{3\left(1+sinx\right)}{cos^2x}-8cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=0\)
Giải PT
a) 4sin (3x + \(\frac{\pi}{3}\)) - 2 = 0
b) 4sin ( 4x + 1) -1 = 0
c) sin ( x + \(\frac{x}{4}\)) -1 = 0
d) 2sin ( 2x + 70o) + 1 = 0
e) sin x . cos ( 2x - 3 ) = 0
f) cos 2x -cos 4x = 0
g) cos ( sin 3x) = 1
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.
giải các pt
a) \(3cos4x-8cos^6x+2cos^2x+3=0\)
b) \(4+3sinx+sin^3x=3cos^2x+cos^6x\)
c) \(2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=cos2x-3\)
d) \(\frac{\sqrt{3}}{cos^2x}-tanx-2\sqrt{3}=sinx\left(1+tanx.tan\frac{x}{2}\right)\)
a/
\(\Leftrightarrow3\left(cos4x+1\right)+2cos^2x\left(1-4cos^4x\right)=0\)
\(\Leftrightarrow3\left(2cos^22x-1+1\right)+2cos^2x\left(1-2cos^2x\right)\left(1+2cos^2x\right)=0\)
\(\Leftrightarrow6cos^22x+\left(1+cos2x\right).\left(-cos2x\right)\left(2+cos2x\right)=0\)
Đặt \(cos2x=a\)
\(\Rightarrow6a^2-a\left(a+1\right)\left(a+2\right)=0\)
\(\Leftrightarrow a\left(-a^2+3a-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\\cos2x=2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow4+3sinx+sin^3x=3\left(1-sin^2x\right)+\left(1-sin^2x\right)^3\)
Đặt \(sinx=a\) ta được:
\(a^3+3a+4=3-3a^2+\left(1-a\right)^3\)
\(\Leftrightarrow a^3+3a^2+3a+1=\left(1-a\right)^3\)
\(\Leftrightarrow\left(a+1\right)^3=\left(1-a\right)^3\)
\(\Leftrightarrow a+1=1-a\)
\(\Leftrightarrow a=0\)
\(\Rightarrow sinx=0\Rightarrow x=k\pi\)
c/
ĐKXĐ: ...
\(\Leftrightarrow2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=2cos^2x-4\)
\(\Leftrightarrow2cos^2x+2cos^2x.tanx.tan\frac{x}{2}=2cos^2x-4\)
\(\Leftrightarrow cos^2x.tanx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}.cos\frac{x}{2}}=-1\)
\(\Leftrightarrow cosx\left(\frac{1-cosx}{2}\right)=-1\)
\(\Leftrightarrow cos^2x-cosx-2=0\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pi+k2\pi\)
1)giải pt a)√2 cos2x-1=0
b) sinx =cos3x
c) cos (x+π/3) +sin(3x+π/4)=0
d)tan 2x = cot (x+π/4)
e) sin x = √3 cos x
f) tan^2(π/3-2x)-3=0
a: \(\Leftrightarrow cos2x=\dfrac{1}{\sqrt{2}}\)
=>2x=pi/4+k2pi hoặc 2x=-pi/4+k2pi
=>x=pi/8+kpi hoặc x=-pi/8+kpi
b: \(\Leftrightarrow sinx=sin\left(\dfrac{pi}{2}-3x\right)\)
=>x=pi/2-3x+k2pi hoặ x=pi/2+3x+k2pi
=>4x=pi/2+k2pi hoặc -2x=pi/2+k2pi
=>x=pi/8+kpi/2 hoặc x=-pi/4-kpi
d: \(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=-sin\left(3x+\dfrac{pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=sin\left(-3x-\dfrac{pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=cos\left(3x+\dfrac{3}{4}pi\right)\)
=>3x+3/4pi=x+pi/3+k2pi hoặc 3x+3/4pi=-x-pi/3+k2pi
=>2x=-5/12pi+k2pi hoặc 4x=-13/12pi+k2pi
=>x=-5/24pi+kpi hoặc x=-13/48pi+kpi/2
e: \(\Leftrightarrow sinx-\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=0\)
=>x-pi/3=kpi
=>x=kpi+pi/3
1) \(sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right).tan^2x-cos^2\frac{x}{2}=0\)
2) \(tanx=sin^2x\left(c-\frac{\pi}{2010}\right)+cos^2\left(2x+\frac{\pi}{2010}\right)+sinx.sin\left(3x+\frac{\pi}{1005}\right)\)
3) \(1+2cosx\left(sinx-1\right)+\sqrt{2}sinx+4cosx.sin^2\frac{x}{2}=0\)
4) \(3cos4x-8cos^6x+2cos4x=3\)
5) \(1+sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)\)
6) \(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-4\sqrt{3}cos^2x.sinx.cos2x\)
7) \(\frac{tan^2x+tanx}{tan^2x+1}=\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{4}\right)\)
8) \(cos^4x+sin^4x+cos\left(x-\frac{\pi}{4}\right).sin\left(3x-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
6.
\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)
\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)
\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải các phương trình sau:
a.\(2sin^3x+4cos^3x=3sinx\)
b.\(3sin^2\frac{x}{2}cos\left(\frac{3\pi}{2}+\frac{x}{2}\right)+3sin^2\frac{x}{2}cos\frac{x}{2}=sin\frac{x}{2}cos^2\frac{x}{2}+sin^2\left(\frac{x}{2}+\frac{\pi}{2}\right)\)
c.\(4sin^3x+3sin^2xcosx-sinx-cos^3x=0\)
d.sin4x-3sin 2xcos2x-4sinxcos3x-3cos4x=0
MỌI NGƯỜI GIÚP MÌNH VỚI MÌNH CẢM ƠN
d.
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^4x\)
\(tan^4x-3tan^2x-4tanx-3=0\)
\(\Leftrightarrow\left(tan^2x+tanx+1\right)\left(tan^2x-tanx-3\right)=0\)
\(\Leftrightarrow tan^2x-tanx-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1-\sqrt{13}}{2}\\tanx=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\frac{1-\sqrt{13}}{2}\right)+k\pi\\x=arctan\left(\frac{1+\sqrt{13}}{2}\right)+k\pi\end{matrix}\right.\)
mọi người giúp hộ mình nhanh với
a.
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(2tan^3x+4=3tanx\left(1+tan^2x\right)\)
\(\Leftrightarrow2tan^3x+4=3tanx+3tan^3x\)
\(\Leftrightarrow tan^3x+3tanx-4=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x+tanx+4\right)=0\)
\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)
Giải PT
a) sin2 x + 2sinx - 3 = 0
b) 2cos x + cos 2x = 0
c) tanx + cotx + 2 = 0
d) sinx + cos2x + 1 = 0
e) tan x + cot 2x = 0
a) TH1: sinx = 1
--> x = pi/2 + k2pi (k nguyên)
TH2: sinx = -3 (loại)
b) 2cosx + cos2x = 0
<=> 2cosx + 2cos^2(x) - 1 = 0
TH1: cosx = (-1 + sqrt(3))/2
TH2: cosx = (-1 - sqrt(3))/2 (loại)
c) ĐKXĐ: x # kpi
Pt <=> tanx + 1/tanx + 2 = 0
--> tanx = -1
--> x = -pi/4 + kpi (k nguyên)
giai cac pt
a) \(sin^3\left(x+\frac{\pi}{4}\right)=\sqrt{2}sinx\)
b) \(cos^3x-sin^3x=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
c) \(\frac{1-tanx}{1+tanx}=1+2sinx\)
d) \(\left(1+tanx\right)sin^2x=3sinx\left(cosx-sinx\right)+3\)
giải các pt
a) \(\left(1+tanx\right)sin^2x=3sinx\left(cosx-sinx\right)+3\)
b) \(6sinx-2cos^3x=\frac{5sin4x.sinx}{2cos2x}\)
c) \(cos^3x=2sinx.sin\left(\frac{\pi}{3}-x\right).sin\left(x+\frac{\pi}{3}\right)\)
d) \(cos2x\left(sinx+cosx\right)-4cos^3x\left(1+sin2x\right)=0\)
a.
ĐKXĐ: \(cosx\ne0\)
Chia 2 vế cho \(cos^2x\) ta được:
\(\left(1+tanx\right).tan^2x=3tanx\left(1-tanx\right)+\frac{3}{cos^2x}\)
\(\Leftrightarrow tan^2x\left(tanx+1\right)=3tanx-3tan^2x+3+3tan^2x\)
\(\Leftrightarrow tan^2x\left(tanx+1\right)-3\left(tanx+1\right)=0\)
\(\Leftrightarrow\left(tan^2x-3\right)\left(tanx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow cos^3x=sinx\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(\Leftrightarrow cos^3x=sinx\left(cos2x-\frac{1}{2}\right)\)
\(\Leftrightarrow cos^3x=2sinx\left(1-2sin^2x-\frac{1}{2}\right)\)
\(\Leftrightarrow cos^3x=sinx\left(\frac{1}{2}-2sin^2x\right)\)
\(\Leftrightarrow2cos^3x=sinx-4sin^3x\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow2=tanx\left(1+tan^2x\right)-4tan^3x\)
\(\Leftrightarrow3tan^3x-tanx+2=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(3tan^2x-3tanx+2\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)
d/
\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(sinx+cosx\right)-4cos^3x\left(sin^2x+cos^2x+2sinx.cosx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)^2-4cos^3x\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left(cosx-sinx-4cos^3x\right)\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx-4cos^3x=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=k\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
Xét \(\left(2\right)\), nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)
\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)
\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)