a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)

\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)

Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)

b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)

\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)

\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)

c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)

\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)

\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)

1.

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

\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Lời giải:
PT $\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=2\sin x\cos x-\sin x$

$\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=\sin x(2\cos x-1)$

$\Leftrightarrow (2\cos x-1)(\sin x+\cos x)=0$

$\Rightarrow 2\cos x=1$ hoặc $\sin x=-\cos x=\cos (\pi -x)=\sin (x-\frac{\pi}{2})$

Đến đây thì đơn giản rồi.

a) <=> 4sinxcosx -(2cos²x-1)=7sinx+2cosx-4

<=> 2cos²x+(2-4sinx)cosx+7sinx-5=0

- sinx=1 => 2cos²x-2cosx+2=0 

pt trên vn

b) <=> 2sinxcosx-1+2sin²x+3sinx-cosx-1=0

<=> cos(2sinx-1)+2sin²x+3sinx-2=0

<=> cosx(2sinx-1)+(2sinx-1)(sinx+2)=0

<=> (2sinx-1)(cosx+sinx+2)=0

<=> sinx=1/2 hoặc cosx+sinx=-2(vn)

<=> x= \(\frac{\pi}{6}+k2\pi\) hoặc \(x=\frac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

a/

\(\Leftrightarrow\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx=sin4x\)

\(\Leftrightarrow sin\left(\frac{\pi}{3}-x\right)=sin4x\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-x+k2\pi\\4x=\frac{2\pi}{3}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)

b/

\(\Leftrightarrow2sinx.cosx+4sinx.cos^2x-2sinx=0\)

\(\Leftrightarrow2sinx\left(cosx+2cos^2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\2cos^2x+cosx-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow2cos4x.sin3x=cos4x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2sin3x=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\3x=\frac{\pi}{6}+k2\pi\\3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

d/

\(\Leftrightarrow6sinx+3cosx+3=sinx-2cosx+3\)

\(\Leftrightarrow sinx+cosx=0\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)

hộ vs ae ơi