Giải pt lượng giác a) sinx- cosx -sin2x +1 =0 b) cosx - sin2x / 2cos^2x - sinx -1 = √3 c) sinx + cosx = 2√2 sinx cosx

a)

$\sin x-\cos x-\sin2x+1=0$

$\Leftrightarrow\sin x-\cos x-2\sin x\cos x+\sin ^x+\cos^2x=0$

$\Leftrightarrow\sin x-\cos x+(\sin x-\cos x)^2=0$

$\Leftrightarrow(\sin x-\cos x)(1+\sin x-\cos x)=0$

$\Leftrightarrow\left[\begin{array}{I}\sin x-\cos x=0\\\sin x-\cos x=-1\end{array}\right.\Leftrightarrow\left[\begin{array}{I}\sin\left({ x-\dfrac{\pi}4}\right)=0\\\sin\left({ x-\dfrac{\pi}4}\right)=-\dfrac1{\sqrt2}\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{I}x-\dfrac{\pi}4=k\pi\\x-\dfrac{\pi}4=-\dfrac{\pi}4+k2\pi\\x-\dfrac{\pi}4=\pi+\dfrac{\pi}4+k2\pi\end{array}\right.\Leftrightarrow\left[\begin{array}{I}x=\dfrac{\pi}4+k\pi\\x=k2\pi\\x=\dfrac{3\pi}2+k2\pi\end{array}\right.$ $(k\in\mathbb Z)$

Vậy phương trình có nghiệm 

$\left\{\begin{array}{I}x=\dfrac{\pi}4+k\pi\\x=k2\pi\\x=\dfrac{3\pi}2+k2\pi\end{array}\right.$ $(k\in\mathbb Z)$

b)

$\dfrac{\cos x-\sin 2x}{2\cos^2x-\sin x-1}=\sqrt3$

Điều kiện $2\cos^2x-\sin x-1\ne0\Leftrightarrow2-2\sin^2x-\sin x-1\ne0$

$\Leftrightarrow\sin x\ne-1$ và $\sin x\ne\dfrac12$

Phương trình suy ra:

$\cos x-\sin2x=\sqrt3.(1+\cos2x)-\sqrt3\sin x-\sqrt3$

$\Leftrightarrow\cos x+\sqrt3\sin x=\sin 2x+\sqrt3\cos 2x$

$\Leftrightarrow\sin\left({x+\dfrac{\pi}6}\right)=\sin\left({2x+\dfrac{\pi}3}\right)$

$\Leftrightarrow\left[\begin{array}{I}x+\dfrac{\pi}6=2x+\dfrac{\pi}3+k2\pi\\x+\dfrac{\pi}6=\pi-2x-\dfrac{\pi}3+k2\pi\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{I}x=-\dfrac{\pi}6+k2\pi\\x=\dfrac{\pi}6+\dfrac{2k\pi}3\end{array}\right.$ $(k\in\mathbb Z)$ (thỏa mãn)

Vậy phương trình có nghiệm

$\left\{\begin{array}{I}x=-\dfrac{\pi}6+k2\pi\\x=\dfrac{\pi}6+\dfrac{2k\pi}3\end{array}\right.$ $(k\in\mathbb Z)$

c)

$\sin x+\cos x=2\sqrt2\sin x\cos x$

$\Leftrightarrow\sqrt2\sin\left({x+\dfrac{\pi}4}\right)=\sqrt2\sin2x$

$\Leftrightarrow\sin\left({x+\dfrac{\pi}4}\right)=\sin2x$

$\Leftrightarrow\left[\begin{array}{I}x+\dfrac{\pi}4=2x+k2\pi\\x+\dfrac{\pi}4=\pi-2x+k2\pi\end{array}\right.\Leftrightarrow\left[\begin{array}{I}x=\dfrac{\pi}4+k2\pi\\x=\dfrac{\pi}4+\dfrac{k2\pi}3\end{array}\right.$

Vậy phương trình có nghiệm

$x=\dfrac{\pi}4+\dfrac{k2\pi}3$ $(k\in\mathbb Z)$.