ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\cosx\ne1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k2\pi\\x\ne k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\frac{1-sin^2x}{1-sinx}=\frac{1-cos^2x}{1-cosx}\)
\(\Leftrightarrow\frac{\left(1-sinx\right)\left(1+sinx\right)}{1-sinx}=\frac{\left(1-cosx\right)\left(1+cosx\right)}{1-cosx}\)
\(\Leftrightarrow1+sinx=1+cosx\)
\(\Leftrightarrow sinx=cosx\Rightarrow x=\frac{\pi}{4}+k\pi\)
Karry Angel đúng ko nè
ĐKXĐ: x≠kπ2x≠kπ2
⇔sinxcosx+cosxsinx=√2(sinx+cosx)⇔sinxcosx+cosxsinx=2(sinx+cosx)
⇔1sinx.cosx=√2(sinx+cosx)⇔(sinx+cosx)sinx.cosx=√22⇔1sinx.cosx=2(sinx+cosx)⇔(sinx+cosx)sinx.cosx=22
Đặt sinx+cosx=asinx+cosx=a (|a|≤√2)(|a|≤2)
⇒a2=1+2sinx.cox⇒sinx.cosx=a2−12⇒a2=1+2sinx.cox⇒sinx.cosx=a2−12 pt trở thành:
(a2−1)a=√2⇔a3−a−√2=0(a2−1)a=2⇔a3−a−2=0
⇔(a−√2)(a2+a√2+1)=0⇒a=√2⇔(a−2)(a2+a2+1)=0⇒a=2
⇒sinx+cosx=√2⇒√2sin(x+π4)=√2⇒sin(x+π4)=1
Đặt |sinx−cosx|=a|sinx−cosx|=a (0≤a≤√20≤a≤2)
⇒1−2sinx.cosx=a2⇒1−sin2x=a2⇒sin2x=1−a2⇒1−2sinx.cosx=a2⇒1−sin2x=a2⇒sin2x=1−a2
Phương trình trở thành:
a+4(1−a2)=1⇔−4a2+a+3=0⇒[a=1a=−34<9(l)a+4(1−a2)=1⇔−4a2+a+3=0⇒[a=1a=−34<9(l)
⇒|sinx−cosx|=1⇔∣∣√2sin(x−π4)∣∣=1⇒|sinx−cosx|=1⇔|2sin(x−π4)|=1
⇔∣∣sin(x−π4)∣∣=√22⇒⎡⎢ ⎢⎣sin(x−π4)=√22sin(x−π4)=−√22⇔|sin(x−π4)|=22⇒[sin(x−π4)=22sin(x−π4)=−22 ⇒...
