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: ...

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)

\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)

\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)

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: \(\left\{{}\begin{matrix}tanx\ne-1\\cosx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left(1+sinx+cos2x\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=cosx\left(1+\frac{sinx}{cosx}\right)\)

\(\Leftrightarrow\left(1+sinx+cos2x\right)\left(sinx+cosx\right)=cosx+sinx\)

\(\Leftrightarrow\left(cosx+sinx\right)\left(sinx+cos2x\right)=0\)

\(\Leftrightarrow sinx+cos2x=0\)

\(\Leftrightarrow-2sin^2x+sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(l\right)\\sinx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

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.

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

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

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

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: \(x\ne k\pi\)

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

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

\(\Leftrightarrow cos2x-sin2x=0\)

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

\(\Leftrightarrow...\)

\(\frac{1+2sinx.cosx}{sin^2x-cos^2x}=\frac{sin^2x+cos^2x+2sinx.cosx}{\left(sinx-cosx\right)\left(sinx+cosx\right)}\)

\(=\frac{\left(sinx+cosx\right)^2}{\left(sinx-cosx\right)\left(sinx+cosx\right)}=\frac{sinx+cosx}{sinx-cosx}\)

\(=\frac{\frac{sinx}{cosx}+\frac{cosx}{cosx}}{\frac{sinx}{cosx}-\frac{cosx}{cosx}}=\frac{tanx+1}{tanx-1}\)

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 \(-2\le a\le2\):

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

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

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

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

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

2.1

a.

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

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

b.

\(cosx-\sqrt{3}sinx=1\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

Cái chỗ biến đổi tương đương cuối cùng bạn làm rõ chút dc ko???

Ném đoạn \(2sin^2x+\left(3\sqrt{2}-2\right)sinx+1\) vào casio mà bấm pt bậc 2 thôi, nó sẽ tách ra biểu thức như cái cuối cùng

Hoặc là tách thế này:

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

\(=2\left[sin^2x-2.\frac{2-3\sqrt{2}}{4}sinx+\left(\frac{2-3\sqrt{2}}{4}\right)^2-\left(\frac{2-3\sqrt{2}}{4}\right)^2\right]+1\)

\(=2\left(sinx-\frac{2-3\sqrt{2}}{4}\right)^2-2\left(\frac{2-3\sqrt{2}}{4}\right)^2+1\)

\(=2\left(sin^2x-\frac{2-3\sqrt{2}}{4}\right)^2+\frac{6\sqrt{2}-7}{4}\)

Với lưu ý \(\frac{6\sqrt{2}-7}{4}>0\) nên biểu thức luôn dương

Câu 1:

ĐKXĐ: \(sin4x\ne0\Rightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\Leftrightarrow4sinx.cos2x+2cos2x=2\)

\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\Leftrightarrow\left(1-2sin^2x\right)\cdot\left(2sinx+1\right)=1\)

\(\Leftrightarrow2sinx-4sin^3x-2sin^2x=0\)

\(\Leftrightarrow sinx\left(-2sin^2x-sinx+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b/

ĐKXĐ: \(sin2x\ne0\Rightarrow x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\frac{sin^4x+cos^4x}{5}=\frac{1}{2}cos2x-\frac{1}{8}\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2=\frac{5}{2}cos2x-\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}sin^22x=\frac{5}{2}cos2x-\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}\left(1-cos^22x\right)=\frac{5}{2}cos2x-\frac{5}{8}\)

\(\Leftrightarrow\frac{1}{2}cos^22x-\frac{5}{2}cos2x+\frac{9}{8}=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2x=\frac{9}{2}>1\left(l\right)\\cos2x=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{\pi}{6}+k\pi\)

ĐKXĐ: \(cosx\ge0\)

\(\Leftrightarrow\left(\sqrt{1-cosx}+\sqrt{cosx}\right)cos2x=sin2x.cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\\sqrt{1-cosx}+\sqrt{cosx}=sin2x\end{matrix}\right.\)

TH1: \(cos2x=0\Rightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

Kết hợp điều kiện \(cosx\ge0\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

TH2: \(\sqrt{1-cosx}+\sqrt{cosx}=sin2x\)

Ta có \(VT=\sqrt{1-cosx}+\sqrt{cosx}\ge\sqrt{1-cosx+cosx}=1\)

\(VP=sin2x\le1\)

\(\Rightarrow VT\ge VP\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}sin2x=1\\\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow ptvn\)