c: Đặt a=sin x+cosx(\(-\sqrt2\le a\le\sqrt2\) )
=>\(a^2=\left(\sin x+cosx\right)^2=\sin^2x+cos^2x+2\cdot\sin x\cdot cosx=1+\sin2x\)
=>\(\sin2x=a^2-1\)
Phương trình ban đầu sẽ trở thành:
\(a^2-1-2\sqrt2\cdot a-5=0\)
=>\(a^2-2\sqrt2a-6=0\)
=>\(\left(a-3\sqrt2\right)\left(a+\sqrt2\right)=0\)
=>\(\left[\begin{array}{l}a=3\sqrt2\left(loại\right)\\ a=-\sqrt2\left(nhận\right)\end{array}\right.\)
=>\(\sin x+cosx=-\sqrt2\)
=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=-\sqrt2\)
=>\(\sin\left(x+\frac{\pi}{4}\right)=-1\)
=>\(x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\)
=>\(x=-\frac34\pi+k2\pi\)
d: \(2\left(\sin x-cosx\right)-3\cdot\sin x\cdot cosx-1=0\)
=>\(2\cdot\left(\sin x-cosx\right)-\frac32\cdot\sin2x-1=0\) (1)
Đặt a=sin x-cosx(\(-\sqrt2\le a\le\sqrt2\) )
=>\(a^2=\left(\sin x-cosx\right)^2=1-\sin2x\)
=>sin 2x=\(1-a^2\)
(1) sẽ trở thành: \(2a-\frac32\cdot\left(1-a^2\right)-1=0\)
=>\(2a-1-\frac32+1,5a^2=0\)
=>\(1,5a^2+2a-2,5=0\)
=>\(3a^2+4a-5=0\)
=>\(a^2+\frac43a-\frac53=0\)
=>\(a^2+\frac43a+\frac49=\frac{19}{9}\)
=>\(\left(a+\frac23\right)^2=\frac{19}{9}\)
=>\(\left[\begin{array}{l}a+\frac23=\frac{\sqrt{19}}{3}\\ a+\frac23=-\frac{\sqrt{19}}{3}\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\frac{\sqrt{19}-2}{3}\left(nhận\right)\\ a=\frac{-\sqrt{19}-2}{3}\left(loại\right)\end{array}\right.\)
=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{19}-2}{3}\)
=>\(\sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{19}-2}{3\sqrt2}=\frac{\sqrt{38}-2\sqrt2}{6}\)
=>\(\left[\begin{array}{l}x-\frac{\pi}{4}=\arcsin\left(\frac{\sqrt{38}-2\sqrt2}{6}\right)+k2\pi\\ x-\frac{\pi}{4}=\pi-\arcsin\left(\frac{\sqrt{38}-2\sqrt2}{6}\right)+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{4}+\arcsin\left(\frac{\sqrt{38}-2\sqrt2}{6}\right)+k2\pi\\ x=\frac{5\pi}{4}-\arcsin\left(\frac{\sqrt{38}-2\sqrt2}{6}\right)+k2\pi\end{array}\right.\)
