a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)

\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)

\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)

\(\Leftrightarrow-cos\left(3x-\dfrac{\pi}{2}\right)+\sqrt{3}cos^3x.\dfrac{cos^2x-3sin^2x}{cos^2x}=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cos3x=-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)-sinx=0\)

\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=k\pi\end{matrix}\right.\)

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

Đối chiếu điều kiện ta được:

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

Lời giải:
ĐKXĐ: \(\left\{\begin{matrix} \cos 2x+1\neq 0\\ \sin x\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2x\neq \pm \pi +2k\pi \\ x\neq n\pi \end{matrix}\right.\) với mọi $k,n\in\mathbb{Z}$

\(\Leftrightarrow \left\{\begin{matrix} x\neq \frac{k}{2}\pi, \text{k nguyên lẻ} \\ x\neq n\pi, \text{n nguyên bất kỳ} \end{matrix}\right.\)

a: \(y=\sqrt{2}sin\left(x+\dfrac{pi}{4}\right)\)

\(-1< =sin\left(x+\dfrac{pi}{4}\right)< =1\)

=>\(-\sqrt{2}< =y< =\sqrt{2}\)

\(y_{min}=-\sqrt{2}\) khi sin(x+pi/4)=-1

=>x+pi/4=-pi/2+k2pi

=>x=-3/4pi+k2pi

\(y_{max}=\sqrt{2}\) khi sin(x+pi/4)=1

=>x+pi/4=pi/2+k2pi

=>x=pi/4+k2pi

b: \(y=sinx\cdot cos\left(\dfrac{pi}{3}\right)+cosx\cdot sin\left(\dfrac{pi}{3}\right)+3\)

\(=sin\left(x+\dfrac{pi}{3}\right)+3\)

-1<=sin(x+pi/3)<=1

=>-1+3<=sin(x+pi/3)+3<=4

=>2<=y<=4

y min=2 khi sin(x+pi/3)=-1

=>x+pi/3=-pi/2+k2pi

=>x=-5/6pi+k2pi

y max=4 khi sin(x+pi/3)=1

=>x+pi/3=pi/2+k2pi

=>x=pi/6+k2pi

c: \(y=2\cdot\left(sin2x\cdot\dfrac{\sqrt{3}}{2}-cos2x\cdot\dfrac{1}{2}\right)\)

\(=2sin\left(2x-\dfrac{pi}{6}\right)\)

-1<=sin(2x-pi/6)<=1

=>-2<=y<=2

y min=-2 khi sin(2x-pi/6)=-1

=>2x-pi/6=-pi/2+k2pi

=>2x=-1/3pi+k2pi

=>x=-1/6pi+kpi

y max=2 khi sin(2x-pi/6)=1

=>2x-pi/6=pi/2+k2pi

=>2x=2/3pi+k2pi

=>x=1/3pi+kpi

a.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)

\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)

b.

\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

c.

\(\Leftrightarrow\dfrac{1}{2}cos6x+\dfrac{1}{2}cos4x=\dfrac{1}{2}cos6x+\dfrac{1}{2}cos2x+4-3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(2cos^22x-1\right)=\dfrac{1}{2}cos2x+\dfrac{5}{2}+\dfrac{3}{2}cos2x\)

\(\Leftrightarrow cos^22x-2cos2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=\pi+k2\pi\)

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

ĐKXĐ:

a. \(x-1\ge0\Rightarrow x\ge1\)

b. \(\left\{{}\begin{matrix}cosx\ne0\\cos2x+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k\pi\\2x\ne\pi+k2\pi\end{matrix}\right.\) \(\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)

c.

\(cosx\ge0\Rightarrow-\dfrac{\pi}{2}+k2\pi\le x\le\dfrac{\pi}{2}+k2\pi\)

a.

\(y'=\dfrac{2-x}{2x^2\sqrt{x-1}}=0\Rightarrow x=2\)

\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)

\(\Rightarrow y_{min}=y\left(1\right)=0\)

\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)

b.

\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]

\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)

c.

\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)

Đặt \(cosx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=-t^2-t+2\)

\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)

\(f\left(-1\right)=2\) ; \(f\left(1\right)=0\) ; \(f\left(-\dfrac{1}{2}\right)=\dfrac{9}{4}\)

\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)

d.

Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)

\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)

\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)

a.

\(y=2\left(1-cos2x\right)-\dfrac{5}{2}sin2x+\dfrac{1}{2}+\dfrac{1}{2}cos2x+10\)

\(=-\dfrac{1}{2}\left(5sin2x+3cos2x\right)+\dfrac{25}{2}\)

\(=-\dfrac{\sqrt{34}}{2}\left(\dfrac{5}{\sqrt{34}}sin2x+\dfrac{3}{\sqrt{34}}cos2x\right)+\dfrac{25}{2}\)

Đặt \(\dfrac{5}{\sqrt{34}}=cosa\)

\(\Rightarrow y=-\dfrac{\sqrt{34}}{2}\left(sin2x.cosa+cos2x.sina\right)+\dfrac{25}{2}\)

\(=-\dfrac{\sqrt{34}}{2}sin\left(2x+a\right)+\dfrac{25}{2}\)

Do \(-1\le sin\left(2x+a\right)\le1\)

\(\Rightarrow\dfrac{25-\sqrt{34}}{2}\le y\le\dfrac{25+\sqrt{34}}{2}\)

b.

\(y=\dfrac{sin^2x-2sin2x+1}{3+sin^2x+2cos^2x}=\dfrac{2sin^2x-4sin2x+2}{6+2\left(sin^2x+cos^2x\right)+2cos^2x}\)

\(=\dfrac{1-cos2x-4sin2x+2}{8+1+cos2x}=\dfrac{3-4sin2x-cos2x}{9+cos2x}\)

\(\Rightarrow9y+y.cos2x=3-4sin2x-cos2x\)

\(\Rightarrow4sin2x+\left(y+1\right)cos2x=3-9y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(4^2+\left(y+1\right)^2\ge\left(3-9y\right)^2\)

\(\Leftrightarrow80y^2-56y-8\le0\)

\(\Rightarrow\dfrac{7-\sqrt{89}}{20}\le y\le\dfrac{7+\sqrt{89}}{20}\)

c.

\(y=2sinx+3cosx+4\)

\(=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}sinx+\dfrac{3}{\sqrt{13}}cosx\right)+4\)

Đặt \(\dfrac{2}{\sqrt{13}}=cosa\)

\(\Rightarrow y=\sqrt{13}\left(sinx.cosa+cosx.sina\right)+4\)

\(=\sqrt{13}sin\left(x+a\right)+4\)

Do \(-1\le sin\left(x+a\right)\le1\)

\(\Rightarrow-\sqrt{13}+4\le y\le\sqrt{13}+4\)