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c/
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sinx=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)
\(\Leftrightarrow sin2x+cos2x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
a/ ĐKXĐ:...
\(\Leftrightarrow\frac{sinx}{cosx}-\frac{\sqrt{2}}{cosx}=1\)
\(\Leftrightarrow sinx-\sqrt{2}=cosx\)
\(\Leftrightarrow sinx-cosx=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{3\pi}{4}+k2\pi\)
b/
ĐKXĐ: ...
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)
\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)
\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)
4.
\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+2sin^2x+3sinx-2=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\frac{\pi}{4}-x=-\frac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)
3.
\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)
\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(pt\Leftrightarrow\dfrac{\left(1-2\sin x\right)\cos x}{1-\sin^2x}=\sqrt{3}\Leftrightarrow\dfrac{1-2\sin x}{\sqrt{3}\cos x}=1\)
\(\Leftrightarrow1-2\sin x=\sqrt{3}\cos x\Leftrightarrow\sqrt{3}\cos x+2\sin x=1\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{\sqrt{7}}\cos x+\dfrac{2}{\sqrt{7}}\sin x=1\)
\(\Leftrightarrow\cos a\cdot\cos x+\sin a\cdot\sin x=1\) với \(a=\sin^{-1}\dfrac{2}{\sqrt{7}}=\cos^{-1}\dfrac{\sqrt{3}}{\sqrt{7}}\)
\(\Leftrightarrow\cos\left(a-x\right)=1\Leftrightarrow a-x=k2\pi\)
\(\Leftrightarrow x=a-k2\pi\Leftrightarrow x=a+m2\pi\left(m\in Z\right)\)
P.s: lâu lâu pick thử bài lượng phát, nguy cơ đúng 80% nhé :)
a/ ĐKXĐ: \(sinx\ne-1\)
\(\Leftrightarrow\left(2sinx+1\right)\left(3cos4x+2sinx\right)+4cos^2x+1=8+8sinx\)
\(\Leftrightarrow6sinx.cos4x+4sin^2x+3cos4x+2sinx+4cos^2x+1=8+8sinx\)
\(\Leftrightarrow6sinx.cos4x+3cos4x-6sinx-3=0\)
\(\Leftrightarrow6sinx\left(cos4x-1\right)+3\left(cos4x-1\right)=0\)
\(\Leftrightarrow\left(6sinx+3\right)\left(cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\cos4x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\1-2sin^22x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1-sin^2x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1+sinx\right)\left(1-sinx\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=k\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}tanx\ne-1\\cosx\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left(1+sinx+cos2x\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=cosx\left(1+\frac{sinx}{cosx}\right)\)
\(\Leftrightarrow\left(1+sinx+cos2x\right)\left(sinx+cosx\right)=cosx+sinx\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(sinx+cos2x\right)=0\)
\(\Leftrightarrow sinx+cos2x=0\)
\(\Leftrightarrow-2sin^2x+sinx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(l\right)\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow2sinx.cosx+2\sqrt{3}cos^2x=\sqrt{3}-2sin5x\)
\(\Leftrightarrow sin2x+\sqrt{3}\left(cos2x+1\right)=\sqrt{3}-2sin5x\)
\(\Leftrightarrow sin2x+\sqrt{3}cos2x=-2sin5x\)
\(\Leftrightarrow\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x=-sin5x\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=sin\left(-5x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=-5x+k2\pi\\2x+\frac{\pi}{3}=\pi+5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{21}+\frac{k2\pi}{7}\\x=-\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow sinx+\sqrt{3}cosx=2sin3x+2sinx\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=-2sin3x\)
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=-sin3x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=sin\left(-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=-3x+k2\pi\\x-\frac{\pi}{3}=\pi+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)
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 \(-2\le a\le2\):
\(a=\frac{3}{a}+1\Leftrightarrow a^2-a-3=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{1+\sqrt{13}}{2}>2\left(l\right)\\a=\frac{1-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Rightarrow2sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{2}\)
\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{4}=sin\alpha\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\alpha+k2\pi\\x+\frac{\pi}{6}=\pi-\alpha+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\)
Giải pt:
\(\left(2sinx-1\right)^2-\left(2sinx-1\right)\left(sinx-\frac{3}{2}\right)=0\)
Giúp với ạ !
\(\Leftrightarrow\left(2sinx-1\right)\left(2sinx-1-sinx+\frac{3}{2}\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
Với \(x\in\left(-\frac{\pi}{4};\frac{\pi}{2}\right)\Rightarrow cosx>0\Rightarrow3cosx+1>0\)
Do đó pt tương đương:
\(2cos2x-1=0\Rightarrow cos2x=\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Pt có 2 nghiệm thuộc khoảng đã cho là \(x=\left\{-\frac{\pi}{6};\frac{\pi}{6}\right\}\)
Với \(cosx=0\) ko phải nghiệm
Với \(cosx\ne0\)
\(\Rightarrow\left(2sin5x-1\right)\left(2cos2x.cosx-cosx\right)=2sinx.cosx\)
\(\Leftrightarrow\left(2sin5x-1\right)\left(cos3x+cosx-cosx\right)=sin2x\)
\(\Leftrightarrow cos3x\left(2sin5x-1\right)=sin2x\)
\(\Leftrightarrow2sin5x.cos3x-cos3x=sin2x\)
\(\Leftrightarrow sin8x+sin2x-cos3x=sin2x\)
\(\Leftrightarrow sin8x=cos3x=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow...\)
Cái em cần là cái "......."
Loại nghiệm mệt lắm sensei