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NV
25 tháng 7 2020

c/

\(a+b+c=1+\sqrt{3}-1-\sqrt{3}=0\)

\(\Rightarrow\) Pt có 2 nghiệm: \(\left[{}\begin{matrix}tanx=1\\tanx=-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

d/ ĐKXĐ: ...

\(\Leftrightarrow cot^22x+3.cot2x+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cot2x=-1\\cot2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{4}+k\pi\\2x=arccot\left(-2\right)+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{1}{2}arccot\left(-2\right)+\frac{k\pi}{2}\end{matrix}\right.\)

NV
25 tháng 7 2020

a/

\(\Leftrightarrow2cos^2x-1+cosx+1=0\)

\(\Leftrightarrow cosx\left(2cosx+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

b/ ĐKXĐ: ...

\(\Leftrightarrow tanx+\frac{1}{tanx}=2\)

\(\Leftrightarrow tan^2x+1=2tanx\)

\(\Leftrightarrow tan^2x-2tanx+1=0\)

\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)

NV
19 tháng 7 2020

c/

\(\Leftrightarrow\sqrt{3}tan\left(\frac{\pi}{9}-2x\right)=-3\)

\(\Leftrightarrow tan\left(\frac{\pi}{9}-2x\right)=-\sqrt{3}\)

\(\Rightarrow\frac{\pi}{9}-2x=-\frac{\pi}{3}+k\pi\)

\(\Rightarrow x=\frac{2\pi}{9}+\frac{k\pi}{2}\)

d/

\(\Leftrightarrow\left[{}\begin{matrix}tanx=5\\tan2x=tan4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\2x=4+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\x=2+\frac{k\pi}{2}\end{matrix}\right.\)

NV
19 tháng 7 2020

a/

ĐKXĐ: ...

\(\Leftrightarrow tanx-8\sqrt{3}=3tanx-6\sqrt{3}\)

\(\Leftrightarrow2tanx=-2\sqrt{3}\)

\(\Rightarrow tanx=-\sqrt{3}\Rightarrow x=-\frac{\pi}{3}+k\pi\)

b/

\(\Leftrightarrow tan2x=-cot\left(\frac{5\pi}{8}\right)\)

\(\Leftrightarrow tan2x=tan\left(\frac{\pi}{2}+\frac{5\pi}{8}\right)\)

\(\Leftrightarrow tan2x=tan\left(\frac{9\pi}{8}\right)\)

\(\Rightarrow2x=\frac{9\pi}{8}+k\pi\Rightarrow x=\frac{9\pi}{16}+\frac{k\pi}{2}\)

NV
19 tháng 7 2020

c/

ĐKXĐ: ...

\(\Leftrightarrow tan2x-2=3\left(2tan2x+1\right)\)

\(\Leftrightarrow5tan2x=-5\)

\(\Rightarrow tan2x=-1\)

\(\Rightarrow2x=-\frac{\pi}{4}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)

d/

ĐKXĐ: ...

\(\Leftrightarrow sinx+\sqrt{3}cosx=3sinx-\sqrt{3}cosx\)

\(\Leftrightarrow2sinx=2\sqrt{3}cosx\)

\(\Rightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi\)

NV
19 tháng 7 2020

a/

\(\Leftrightarrow tanx=-tan\left(\frac{2\pi}{3}-3x\right)\)

\(\Leftrightarrow tanx=tan\left(3x-\frac{2\pi}{3}\right)\)

\(\Rightarrow x=3x-\frac{2\pi}{3}+k\pi\)

\(\Rightarrow x=\frac{\pi}{3}+\frac{k\pi}{2}\)

b/

\(tan\left(2x-15^0\right)=tanx\)

\(\Rightarrow2x-15^0=x+k180^0\)

\(\Rightarrow x=15^0+k180^0\)

NV
26 tháng 7 2020

e/

ĐKXĐ: ...

\(\Leftrightarrow\frac{1}{cos^2x}\left(9-13cosx\right)+4=0\)

\(\Leftrightarrow\frac{9}{cos^2x}-\frac{13}{cosx}+4=0\)

Đặt \(\frac{1}{cosx}=t\)

\(\Rightarrow9t^2-13t+4=0\)

\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{4}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=1\\\frac{1}{cosx}=\frac{4}{9}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=k2\pi\)

NV
26 tháng 7 2020

d/

\(\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0\)

\(\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0\)

\(\Leftrightarrow-2sin^22x+sin2x+1=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)

NV
25 tháng 7 2020

d/

ĐKXĐ: ...

\(\Leftrightarrow tanx-1+cos2x=0\)

\(\Leftrightarrow\frac{sinx}{cosx}-1-\left(sin^2x-cos^2x\right)=0\)

\(\Leftrightarrow\frac{sinx-cosx}{cosx}-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{cosx}-sinx-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\\frac{1}{cosx}-sinx-cosx=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)

\(\Rightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)

\(\left(2\right)\Leftrightarrow1-sinx.cosx-cos^2x=0\)

\(\Leftrightarrow sin^2x-sinx.cosx=0\)

\(\Leftrightarrow sinx\left(sinx-cosx\right)=0\)

\(\Leftrightarrow sinx=0\Rightarrow x=k\pi\)

NV
25 tháng 7 2020

c/

\(\Leftrightarrow sinx.cos2x-sinx+1-cos2x=0\)

\(\Leftrightarrow sinx\left(cos2x-1\right)-\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\cos2x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\2x=k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=k\pi\end{matrix}\right.\)

28 tháng 7 2019
https://i.imgur.com/mVqlQRs.jpg
28 tháng 7 2019
https://i.imgur.com/tXKeJRL.jpg
NV
25 tháng 7 2020

b/

\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cosx+1-cos^2x+2cos^2x-1=\frac{1}{2}\)

\(\Leftrightarrow cos^2x+\frac{1}{2}cosx=0\)

\(\Leftrightarrow cosx\left(cosx+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\left(\frac{sinx}{cosx}+\frac{cosx}{sinx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{sinx.cosx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{2}{sin2x}\right)^2+\frac{3}{sin2x}-7=0\)

Đặt \(\frac{1}{sin2x}=a\Rightarrow4a^2+3a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{4}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=arcsin\left(-\frac{4}{7}\right)+k2\pi\\2x=\pi-arcsin\left(-\frac{4}{7}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\end{matrix}\right.\)

NV
25 tháng 7 2020

a/

\(\Leftrightarrow2cos2x.cosx+\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos^22x=0\)

\(\Leftrightarrow cos2x\left(2cosx+cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\left(1\right)\\2cosx+cos2x=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(\left(2\right)\Leftrightarrow2cosx+2cos^2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}-1}{2}\\cosx=\frac{-\sqrt{3}-1}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\)

16 tháng 7 2020

\(\frac{tanx-1}{tanx+1}+cot2x=0\\ \Leftrightarrow cot2x-\frac{1-tanx\cdot tan\frac{\pi}{4}}{tanx+tan\frac{\pi}{4}}=0\\ \Leftrightarrow cot2x-cot\left(x+\frac{\pi}{4}\right)=0\)

NV
16 tháng 7 2020

d/

ĐKXĐ: \(\left\{{}\begin{matrix}sin2x\ne0\\tanx\ne-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}+cot2x=0\\3tanx-\sqrt{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}-\frac{tan^2x-1}{2tanx}=0\\tanx=\frac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(tanx-1\right)\left(\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}\right)=0\left(1\right)\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Xét (1): \(\Leftrightarrow\left[{}\begin{matrix}tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\\\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}=0\left(2\right)\end{matrix}\right.\)

Xét (2)

\(\Leftrightarrow\left(tanx+1\right)^2-2tanx=0\)

\(\Leftrightarrow tan^2x+1=0\left(vn\right)\)

NV
13 tháng 7 2020

10. ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(2cos2x+tanx=\frac{4}{5}\)

\(\Leftrightarrow4cos^2x-2+tanx=\frac{4}{5}\)

\(\Leftrightarrow\frac{4}{1+tan^2x}+tanx-\frac{14}{5}=0\)

Đặt \(tanx=t\)

\(\Rightarrow\frac{20}{1+t^2}+5t-14=0\)

\(\Leftrightarrow5t^3-14t^2+5t+6=0\)

\(\Leftrightarrow\left(t-2\right)\left(5t^2-4t-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{2+\sqrt{19}}{5}\\t=\frac{2-\sqrt{19}}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=2=tana\\tanx=\frac{2+\sqrt{19}}{5}=tanb\\tanx=\frac{2-\sqrt{19}}{5}=tanc\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\\x=c+k\pi\end{matrix}\right.\)

NV
13 tháng 7 2020

9.

\(\Leftrightarrow cos2x-3cosx=2\left(cosx+1\right)\)

\(\Leftrightarrow2cos^2x-1-3cosx=2cosx+2\)

\(\Leftrightarrow2cos^2x-5cosx-3=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=3\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)

19 tháng 9 2020

Bài 3:  Một số phương trình lượng giác thường gặp

NV
19 tháng 9 2020

a.

\(\Leftrightarrow2sin\frac{17\pi}{30}cos\left(3x-\frac{7\pi}{30}\right)=\sqrt{3}\)

\(\Leftrightarrow cos\left(3x-\frac{7\pi}{30}\right)=\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}\)

Đặt \(\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cos\left(3x-\frac{7\pi}{30}\right)=cosa\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{7\pi}{30}=a+k2\pi\\3x-\frac{7\pi}{30}=-a+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{90}+\frac{a}{3}+\frac{k2\pi}{3}\\x=\frac{7\pi}{30}-\frac{a}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)

Chắc bạn ghi sai đề, con số \(\frac{4\pi}{3}\) sẽ hợp lý hơn con số \(\frac{4\pi}{5}\) rất nhiều