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bạn ơi, cho mình hỏi là tại sao từ bước 2 xuống bước 3, tử sin22x-2 lại đổi thành 2-sin22x vậy ạ
Nhân cả tử và mẫu với -1 thôi bạn
\(=\frac{2-sin^22x}{4cos^2x\left(1-sin^2x\right)}=\frac{2-sin^2x}{4cos^2x.cos^2x}\)
a.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)
\(\Leftrightarrow1-sin^2x=0\)
\(\Leftrightarrow cos^2x=0\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
b.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
\(sin^2x+sin^22x=1\)
\(\Leftrightarrow2sin^2x-1+2sin^22x-2=-1\)
\(\Leftrightarrow-cos2x-2cos^22x+1=0\)
\(\Leftrightarrow\left(cos2x+1\right)\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi+k2\pi\\2x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\left(tanx+cotx\right)^2=\frac{4+sin4x}{sin^22x}+2\)
\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{\frac{1}{2}sin2x}\right)^2=\frac{4+sin4x+2sin^22x}{sin^22x}\)
\(\Leftrightarrow\frac{4}{sin^22x}=\frac{4+sin4x+2sin^22x}{sin^22x}\)
\(\Leftrightarrow2sin^22x+sin4x=0\)
\(\Leftrightarrow1-cos4x+sin4x=0\)
\(\Leftrightarrow\sqrt{2}cos\left(4x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\left(l\right)\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow\frac{\cos^2x-4\sin^2x.\cos^2x}{4\cos^2x}=\frac{1}{2}\left(\cos\frac{\pi}{3}-\cos2x\right)\)
\(\Leftrightarrow1-4\sin^2x=2\left(\frac{1}{2}-\cos2x\right)\)
\(\Leftrightarrow1-4\sin^2x=1-2\cos2x\)
\(\Leftrightarrow2\sin^2x=\cos2x\)
\(\Leftrightarrow1-\cos2x=\cos2x\)
\(\Leftrightarrow\cos2x=\frac{1}{2}\Leftrightarrow x=\pm\frac{\pi}{6}+k\pi,k\in Z\) thỏa mãn điều kiện
a)\(pt\Leftrightarrow\frac{1-cos8x}{2}+\frac{1-cos6x}{2}=\frac{1-cos4x}{2}+\frac{1-cos2x}{2}\)
\(\Leftrightarrow cos2x+cos4x=cos6x+cos8x\)
\(\Leftrightarrow2cos3x\cdot cosx=2cos7x\cdot cosx\)
\(\Leftrightarrow2cos\left(cos3x-cos7x\right)=0\)
\(\Leftrightarrow2cosx\cdot\left(-2\right)\cdot sin5x\cdot sin\left(-2x\right)=0\)
\(\Leftrightarrow cosx\cdot sin2x\cdot sin5x=0\)
\(\Leftrightarrow sin2x\cdot sin5x=0\)(do sin2x=0 <=>2sinx*cosx=0 gồm th cosx=0 r`)
\(\Leftrightarrow\left[\begin{array}{nghiempt}sin2x=0\\sin5x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{k\pi}{2}\\x=\frac{k\pi}{5}\end{array}\right.\)\(\left(k\in Z\right)\)
b)\(pt\Leftrightarrow1-cos2x+1-cos4x=1+cos6x+1+cos8x\)
\(\Leftrightarrow cos2x+cos8x+cos4x+cos6x=0\)
\(\Leftrightarrow cos10x\cdot cos6x+cos10x\cdot cos2x=0\)
\(\Leftrightarrow cos10x\left(cos6x+cos2x\right)=0\)
\(\Leftrightarrow cos10x\cdot cos8x\cdot cos4x=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}cos10x=0\\cos8x=0\\cos4x=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{16}+\frac{k\pi}{8}\\x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
b.
\(\Leftrightarrow2sin^2x+4sinx=3\left(1-sin^2x\right)\)
\(\Leftrightarrow5sin^2x+4sinx-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{-2-\sqrt{19}}{5}\left(l\right)\\sinx=\frac{-2+\sqrt{19}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{-2+\sqrt{19}}{5}\right)+k2\pi\\x=\pi-arcsin\left(\frac{-2+\sqrt{19}}{5}\right)+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow sinx\left(sin^2x+3sinx+2\right)=0\)
\(\Leftrightarrow sinx\left(sinx+1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
a.
\(1-cos^22x-\left(\frac{1-cos2x}{2}\right)=\frac{1}{2}\)
\(\Leftrightarrow2cos^22x-cos2x=0\)
\(\Leftrightarrow cos2x\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=\frac{1}{2}\\\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4sin^2x}+1=2tan^2x\)
\(\Leftrightarrow\frac{4sin^2x\left(cos^2x-1\right)}{4sin^2x\left(cos^2x+1\right)}+1=\frac{2sin^2x}{cos^2x}\)
\(\Leftrightarrow\frac{cos^2x}{cos^2x+1}=\frac{1-cos^2x}{cos^2x}\)
Đặt \(cos^2x=t\Rightarrow0< t< 1\)
\(\Rightarrow\frac{t}{t+1}=\frac{1-t}{t}\Leftrightarrow t^2=1-t^2\Leftrightarrow t^2=\frac{1}{2}\)
\(\Leftrightarrow t=\frac{\sqrt{2}}{2}\Leftrightarrow cos^2x=\frac{\sqrt{2}}{2}\)