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24 tháng 8 2021

1.

\(8sinx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\)

\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}}{2cosx}+\dfrac{1}{2sinx}\)

\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}sinx+cosx}{sin2x}\)

\(\Leftrightarrow4sinx.sin2x=\sqrt{3}sinx+cosx\)

\(\Leftrightarrow2cosx-2cos3x=\sqrt{3}sinx+cosx\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=2cos3x\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos3x\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=\pm3x+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}-k\pi\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)

24 tháng 8 2021

2.

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(sinx+\sqrt{3}cosx=\dfrac{1}{cosx}\)

\(\Leftrightarrow2sinx.cosx+2\sqrt{3}cos^2x-\sqrt{3}=2-\sqrt{3}\)

\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=1-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{2-\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\\2x+\dfrac{\pi}{3}=\pi-arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\\x=\dfrac{\pi}{3}-\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\end{matrix}\right.\)

NV
22 tháng 9 2019

a/ ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k2\pi\\x\ne-\frac{\pi}{6}+k2\pi\\x\ne\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(1+sinx-2sin^2x\right)\)

\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(cos2x+sinx\right)\)

\(\Leftrightarrow\sqrt{3}sinx-cosx=sin2x+\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=sin\left(2x+\frac{\pi}{6}\right)\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(cosx+\sqrt{3}sinx\ne0\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)\ne0\Rightarrow...\)

Đặt \(cosx+\sqrt{3}sinx=2sin\left(x+\frac{\pi}{6}\right)=a\) với \(-2\le a\le2\):

\(a=\frac{3}{a}+1\Leftrightarrow a^2-a-3=0\)

\(\Rightarrow\left[{}\begin{matrix}a=\frac{1+\sqrt{13}}{2}>2\left(l\right)\\a=\frac{1-\sqrt{13}}{2}\end{matrix}\right.\)

\(\Rightarrow2sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{2}\)

\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{4}=sin\alpha\)

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

NV
22 tháng 7 2020

d/

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)

\(\Leftrightarrow tanx=\sqrt{3}\)

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

NV
22 tháng 7 2020

c/

ĐKXĐ: \(sin2x\ne0\)

\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)

\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)

\(\Leftrightarrow1-cosx=sin^2x\)

\(\Leftrightarrow1-cosx=1-cos^2x\)

\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

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

NV
27 tháng 9 2020

ĐKXĐ: \(sinx\ne\frac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{6}+k2\pi\\x\ne\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Pt tương đương:

\(cosx=\sqrt{3}sinx\Leftrightarrow tanx=\frac{1}{\sqrt{3}}\)

\(\Leftrightarrow x=\frac{\pi}{6}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\frac{7\pi}{6}+k2\pi\)

NV
28 tháng 10 2020

d.

\(\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)

e.

\(\Leftrightarrow cosx.cos\left(\frac{\pi}{12}\right)-sinx.sin\left(\frac{\pi}{12}\right)=\frac{1}{2}\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{12}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{12}=\frac{\pi}{3}+k2\pi\\x+\frac{\pi}{12}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

NV
28 tháng 10 2020

2.a.

ĐKXĐ: ...

\(\sqrt{3}tanx-\frac{6}{tanx}+2\sqrt{3}-3=0\)

\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-2\\tanx=\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-2\right)+k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ne k\pi\)

\(1-sin2x=2sin^2x\)

\(\Leftrightarrow1-2sin^2x-sin2x=0\)

\(\Leftrightarrow cos2x-sin2x=0\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow...\)