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1.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3=1+\frac{1-t^2}{2}\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)
\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)
\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)
(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)
\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)
b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx+\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)
\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)
\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)
(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)
\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)
\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)
d/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)
\(\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
c/
ĐKXĐ: \(sin2x\ne0\)
\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)
\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)
\(\Leftrightarrow1-cosx=sin^2x\)
\(\Leftrightarrow1-cosx=1-cos^2x\)
\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)
Đểphương trình có nghĩa thì \(\left\{{}\begin{matrix}x\ne\dfrac{\Pi}{2}+k\Pi\\\tan x< >\sin x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\Pi}{2}+k\Pi\\\dfrac{\sin x}{\cos x}\ne\sin x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\Pi}{2}+k\Pi\\\sin x\left(\cos x-1\right)< >0\end{matrix}\right.\Leftrightarrow x\notin\left\{\dfrac{\Pi}{2}+k\Pi;k\Pi;k2\Pi\right\}\)