a,
$2(\sin x+\cos x)-2.2\sin x\cos x-1=0$
Đặt $t=\sin x+\cos x=\sqrt2.\sin\left(x+\dfrac{\pi}{4}\right)$ ($t\in [-\sqrt2;\sqrt2]$)
$\to t^2-1=2\sin x\cos x$
PT trở thành:
$2t-2(t^2-1)-1=0$
$\to -2t^2+2t+1=0$
$\to t=\dfrac{1\pm\sqrt3}{2}$
$\to \sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1\pm\sqrt3}{2\sqrt2}$
$\to \left[ \begin{array}{l}x=\dfrac{-\pi}{4}+\arcsin\dfrac{1\pm\sqrt3}{2\sqrt2}+k2\pi \\x=\dfrac{3\pi}{4}-\arcsin\dfrac{1\pm\sqrt3}{2\sqrt2}+k2\pi \end{array} \right.$
b,
$2\sin x\cos x-3(\sin x+\cos x)+3=0$
Đặt $t=\sin x+\cos x=\sqrt2.\sin\left(x+\dfrac{\pi}{4}\right)$ ($t\in [-\sqrt2;\sqrt2]$)
$\to t^2-1=2\sin x\cos x$
PT trở thành:
$t^2-1-3t+3=0$
$\to t=1$
$\to \sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}$
$\to \left[ \begin{array}{l}x=k2\pi \\x= \dfrac{\pi}{2}+k2\pi \end{array} \right.$
c,
$5(\sin x-\cos x)-5-2\sin x\cos x=0$
Đặt $t=\sin x-\cos x=\sqrt2.\sin\left(x-\dfrac{\pi}{4}\right)$ ($t\in [-\sqrt2;\sqrt2]$)
$\to t^2-1=-2\sin x\cos x$
PT trở thành:
$5t-5+t^2-1=0$
$\to t=1$
$\to \sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi \end{array} \right.$
d,
$\sin x-\cos x-2.2\sin x\cos x-1=0$
Đặt $t=\sin x-\cos x=\sqrt2.\sin\left(x-\dfrac{\pi}{4}\right)$ ($t\in [-\sqrt2;\sqrt2]$)
$\to t^2-1=-2\sin x\cos x$
PT trở thành:
$t+2(t^2-1)-1=0$
$\to 2t^2+t-3=0$
$\to t=1$
$\to \sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi \end{array} \right.$
e,
$3(\sin x+\cos x)-2.2\sin x\cos x=0$
Đặt $t=\sin x+\cos x=\sqrt2.\sin\left(x+\dfrac{\pi}{4}\right)$ ($t\in [-\sqrt2;\sqrt2]$)
$\to t^2-1=2\sin x\cos x$
PT trở thành:
$3t-2(t^2-1)=0$
$\to -2t^2+3t+2=0$
$\to t=\dfrac{-1}{2}$
$\to \sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{-1}{2\sqrt2}$
$\to \left[ \begin{array}{l}x=\dfrac{-\pi}{4}+arcsin\dfrac{-1}{2\sqrt2}+k2\pi \\x=\dfrac{3\pi}{4}-\arcsin\dfrac{-1}{2\sqrt2}+k2\pi \end{array} \right.$
f,
$\sin x+\cos x-2\sin x\cos x=0$
Đặt $t=\sin x+\cos x=\sqrt2.\sin\left(x+\dfrac{\pi}{4}\right)$ ($t\in [-\sqrt2;\sqrt2]$)
$\to t^2-1=2\sin x\cos x$
PT trở thành: $t-t^2+1=0$
$\to t=\dfrac{1-\sqrt5}{2}$
$\to \sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1-\sqrt5}{2\sqrt2}$
$\to \left[ \begin{array}{l}x=\dfrac{-\pi}{4}+\arcsin\dfrac{1-\sqrt5}{2\sqrt2}+k2\pi \\x=\dfrac{3\pi}{4}-\arcsin\dfrac{1-\sqrt5}{2\sqrt2}+k2\pi \end{array} \right.$
