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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) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)
\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)
\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)
2/
\(\Leftrightarrow3sinx-4sin^3x-\sqrt{3}cosx=2sinx\)
\(\Leftrightarrow4sin^3x-sinx+\sqrt{3}cosx=0\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow4tan^3x-tanx\left(1+tan^2x\right)+\sqrt{3}\left(1+tan^2x\right)=0\)
\(\Leftrightarrow3tan^3x+\sqrt{3}tan^2x-tanx+\sqrt{3}=0\)
Bạn xem lại đề, pt bậc 3 này ko giải được (nghiệm rất xấu)
1.
\(\Leftrightarrow\sqrt{3}cos^2x-\sqrt{3}+cos^2x+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)
\(\Leftrightarrow-\sqrt{3}sin^2x+cosx+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+\sqrt{3}sinx\right)-\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+\sqrt{3}sinx-1\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx-\frac{1}{2}\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\left[sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)
\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)
\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)
\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)
\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)
\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)
ĐKXĐ: \(x\ne-\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow\frac{\left(1-sin^2x\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1+sinx\right)\)
\(\Leftrightarrow\frac{\left(1+sinx\right)\left(1-sinx\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1+sinx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\\\frac{\left(1-sinx\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cosx-1-sinx.cosx+sinx=2sinx+2cosx\)
\(\Leftrightarrow sinx+cosx+sinx.cosx+1=0\)
\(\Leftrightarrow\left(sinx+1\right)\left(cosx+1\right)=0\)
\(\Leftrightarrow...\)
\(cos\left(2x-18^o\right).tan50^0+sin\left(2x-18^o\right)=\dfrac{1}{2cos130^0}\)
⇔\(cos\left(2x-18^o\right).sin50^0+sin\left(2x-18^o\right).cos50^0=\dfrac{cos50^0}{2cos130^0}\)
(Nhân cả 2 vế với cos500)
⇔ sin (500 + 2x - 180) = \(-\dfrac{1}{2}\)
⇔ \(\left[{}\begin{matrix}2x+32^0=-30^0+k.360^0\\2x+32^0=210^0+k.360^0\end{matrix}\right.\) với k là số nguyên