\(tan\left(\frac{\Pi}{3}-3x\right)+tan2x+tanx=\sqrt{3}\)
\(4sin^2\left(x+\frac{\Pi}{6}\right)+sin2x=1\)
giải phương trình sau:
a,\(\frac{sin2x+2cosx-sinx-1}{tanx+\sqrt{3}}=0\)
b,\(\frac{\left(1+sinx+cos2x\right)sinx\left(x+\frac{\pi}{4}\right)}{1+tanx}=\frac{1}{\sqrt{2}}cosx\)
c,\(\frac{\left(1-sin2x\right)cosx}{\left(1+sin2x\right)\left(1-sinx\right)}=\sqrt{3}\)
d,\(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
Giair các pt lượng giác sau:
1) \(sin\left(x-\frac{\pi}{4}\right)\left(2cos+\sqrt{2}\right)tan2x=0\)
2) \(tan2x.sinx+3\left(sin-\sqrt{3}tan2x\right)-3\sqrt{3}=0\)
3) \(\frac{cos2x}{sin\left(x+\frac{3\pi}{4}\right)}=\frac{sin\left(x+\frac{3\pi}{4}\right)}{cos2x}\)
4) \(\left(\frac{tanx-1}{tanx+1}+cot2x\right)\left(3tan-\sqrt{3}\right)=0;0< x< \pi\)
a/ ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\2cosx+\sqrt{2}=0\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\cosx=cos\left(\frac{3\pi}{4}\right)\\2x=k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=\frac{3\pi}{4}+k2\pi\left(l\right)\\x=-\frac{3\pi}{4}+k2\pi\left(l\right)\\x=\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x=\frac{k\pi}{2}\)
b/
ĐKXĐ: \(x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\Leftrightarrow tan2x.sinx+3sinx-\sqrt{3}tan2x-3\sqrt{3}=0\)
\(\Leftrightarrow sinx\left(tan2x+3\right)-\sqrt{3}\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\right)\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\sqrt{3}>1\left(vn\right)\\tan2x=-3\end{matrix}\right.\)
\(\Rightarrow2x=arctan\left(-3\right)+k\pi\)
\(\Rightarrow x=\frac{arctan\left(-2\right)}{2}+\frac{k\pi}{2}\)
c/
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x+\frac{3\pi}{4}\right)\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\frac{3\pi}{4}\ne k\pi\\2x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne-\frac{3\pi}{4}+k\pi\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(cos^22x=sin^2\left(x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{3\pi}{2}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{3\pi}{2}\right)=cos\left(2x+\frac{\pi}{2}\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=2x+\frac{\pi}{2}+k2\pi\\4x=-2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)
\(tan\left(\frac{\Pi}{3}-3x\right)+tan2x+tanx=\sqrt{3}\)
\(tan\left(\frac{\Pi}{3}-3x\right)+tan2x+tanx=\sqrt{3}\)
\(tana+tanb=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}\)
\(tana-tanb=\frac{sina.cosb-cosa.sinb}{cosa.cosb}=\frac{sin\left(a-b\right)}{cosa.cosb}\)
\(tan\left(\frac{\pi}{3}-3x\right)-tan\left(\frac{\pi}{3}\right)+tan2x+tanx=0\)
\(\Leftrightarrow\frac{-sin3x}{cos\left(\frac{\pi}{3}-3x\right).cos\left(\frac{\pi}{3}\right)}+\frac{sin3x}{cosx.cos2x}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cosx.cos2x=\frac{1}{2}cos\left(\frac{\pi}{3}-3x\right)\end{matrix}\right.\)
Pt dưới \(\Leftrightarrow cos3x+cosx=cos\left(\frac{\pi}{3}-3x\right)\)
\(\Leftrightarrow cos3x-cos\left(\frac{\pi}{3}-3x\right)+cosx=0\)
\(\Leftrightarrow-2sin\left(\frac{\pi}{6}\right).sin\left(3x-\frac{\pi}{6}\right)+cosx=0\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=-cosx=sin\left(x-\frac{\pi}{2}\right)\)
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 pt
a) \(cos^2\left(\frac{\pi}{3}+x\right)+4cos\left(\frac{\pi}{6}-x\right)=4\)
b) \(5cos\left(2x+\frac{\pi}{3}\right)=4sin\left(\frac{5\pi}{6}-x\right)-9\)
c) \(2sin^2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
d) \(2sin^2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
a/
Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)
Pt trở thành:
\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)
\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)
\(\Leftrightarrow cos^2a+4sina-4=0\)
\(\Leftrightarrow1-sin^2a+4sina-4=0\)
\(\Leftrightarrow-sin^2a+4sina-3=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=1\)
\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)
b/
Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)
Pt trở thành:
\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)
\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)
\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)
\(\Leftrightarrow10sin^2a+4sina-14=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)
\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
giải các pt
a) \(\frac{tanx-8\sqrt{3}}{tanx-2\sqrt{3}}=3\)
b) \(tan2x+cot\frac{5\pi}{8}=0\)
c) \(\sqrt{3}tan\left(\frac{\pi}{9}-2x\right)+3=0\)
d) \(\left(tanx-5\right)\left(tan4-tan2x\right)=0\)
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/
\(\Leftrightarrow tan2x=-cot\left(\frac{5\pi}{8}\right)\)
\(\Leftrightarrow tan2x=tan\left(\frac{\pi}{2}+\frac{5\pi}{8}\right)\)
\(\Leftrightarrow tan2x=tan\left(\frac{9\pi}{8}\right)\)
\(\Rightarrow2x=\frac{9\pi}{8}+k\pi\Rightarrow x=\frac{9\pi}{16}+\frac{k\pi}{2}\)
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/
\(\Leftrightarrow\left[{}\begin{matrix}tanx=5\\tan2x=tan4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\2x=4+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\x=2+\frac{k\pi}{2}\end{matrix}\right.\)
giải pt
a) \(tan\left(x+\frac{\pi}{3}\right)=cot\left(3x-\frac{\pi}{6}\right)\)
b) \(\left(\sqrt{3}-\frac{3}{tanx}\right).tanx=0\)
a/
\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=tan\left(\frac{2\pi}{3}-3x\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\frac{2\pi}{3}-3x+k\pi\)
\(\Rightarrow4x=\frac{\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{\pi}{12}+\frac{k\pi}{4}\)
b/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{3}-\frac{3}{tanx}=0\)
\(\Leftrightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi\)
giải pt
a) \(tanx.tan\frac{\pi}{9}=1+tan\frac{\pi}{9}.tan\frac{\pi}{90}+tanx.tan\frac{\pi}{90};\left(-2\pi< x< 2\pi\right)\)
b) \(tan^22x+\frac{1}{cos^22x}=7;\left(0< x< 360^0\right)\)
c) \(tan^3x+\frac{1}{cos^2x}+4\sqrt{3}\left(1+tanx\right)=8+7tanx;\left(-\pi< x< \pi\right)\)
a/ \(\Leftrightarrow tanx.tan\frac{\pi}{9}-1=tan\frac{\pi}{90}\left(tanx+tan\frac{\pi}{9}\right)\)
\(\Leftrightarrow\frac{tanx+tan\frac{\pi}{9}}{1-tanx.tan\frac{\pi}{9}}=-\frac{1}{tan\frac{\pi}{90}}\)
\(\Leftrightarrow tan\left(x+\frac{\pi}{9}\right)=tan\left(\frac{23\pi}{45}\right)\)
\(\Rightarrow x+\frac{\pi}{9}=\frac{23\pi}{45}+k\pi\)
\(\Rightarrow x=\frac{2\pi}{5}+k\pi\)
Do \(-2\pi< x< 2\pi\Rightarrow-2\pi< \frac{2\pi}{5}+k\pi< 2\pi\)
\(\Rightarrow k=\left\{-2;-1;0;1;2\right\}\)
\(\Rightarrow x=\left\{-\frac{8\pi}{5};-\frac{3\pi}{5};\frac{2\pi}{5};\frac{7\pi}{5};\frac{12\pi}{5}\right\}\)
b/
ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow tan^22x+1+tan^22x=7\)
\(\Leftrightarrow tan^22x=3\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=tan60^0\\tan2x=tan\left(-60^0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=60^0+k180^0\\2x=-60^0+k180^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-30^0+k180^0\end{matrix}\right.\)
Bạn tự tìm nghiệm thuộc khoảng đã cho nhé
c/ ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow tan^3x+1+tan^2x+4\sqrt{3}\left(1+tanx\right)=8+7tanx\)
\(\Leftrightarrow tan^2x\left(1+tanx\right)+\left(4\sqrt{3}-7\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left(tan^2x-7+4\sqrt{3}\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tan^2x=7-4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=tan\left(-\frac{\pi}{4}\right)\\tanx=tan\left(\frac{\pi}{12}\right)\\tanx=tan\left(-\frac{\pi}{12}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{12}+k\pi\\x=-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
Bạn tự tìm x thuộc khoảng đã cho