d: cos^2x=1

=>sin^2x=0

=>sin x=0

=>x=kpi

a: =>sin 4x=cos(x+pi/6)

=>sin 4x=sin(pi/2-x-pi/6)

=>sin 4x=sin(pi/3-x)

=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi

=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3

b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi

=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi

c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi

=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2

\(\Leftrightarrow cos6x-cos8x+2\left(1-cos4x\right)^2+\sqrt{3}sin6x=4-4cos4x\)

\(\Leftrightarrow cos6x-cos8x+2\left(1+cos^24x-2cos4x\right)+\sqrt{3}sin6x=4-4cos4x\)

\(\Leftrightarrow cos6x-cos8x+cos8x+3-4cos4x+\sqrt{3}sin6x=4-4cos4x\)

\(\Leftrightarrow cos6x+\sqrt{3}sin6x=1\)

\(\Leftrightarrow cos\left(6x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow...\)

1.

\(2cos4x-3=0\)

\(\Leftrightarrow cos4x=\dfrac{3}{2}\)

Mà \(cos4x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

2.

\(cos5x+2=0\)

\(\Leftrightarrow cos5x=-2\)

Mà \(cos5x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

3.

\(cos2x+0,7=0\)

\(\Leftrightarrow cos2x=-\dfrac{7}{10}\)

\(\Leftrightarrow2x=\pm arccos\left(-\dfrac{7}{10}\right)+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{arccos\left(-\dfrac{7}{10}\right)}{2}+k\pi\)

4.

\(cos^22x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-\dfrac{1}{2}\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

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

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

Ta có:

\(\dfrac{1}{2}\left(cos4x+cos2x\right)-cos4x=2-4sin^2\left(\dfrac{\pi}{4}-\dfrac{3x}{2}\right)\)

<=> \(\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x-cos4x=2\left(1-2sin^2\left(\dfrac{n}{4}-\dfrac{3x}{2}\right)\right)\)

<=> \(\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x=2cos\left(\dfrac{n}{2}-3x\right)\)

<=> \(\dfrac{1}{2}\left(cos2x-cos4x\right)=2sin3x\)

<=> \(\dfrac{1}{2}\left(-2\right)sin3xsin\left(-x\right)=2sin3x\)

<=>\(\dfrac{1}{2}\left(-2\right)\left(-1\right)sin3xsinx=2sin3x\)

<=>\(sin3xsinx=2sin3x\)

<=> \(sin3xsinx-2sin3x=0\)

<=>\(sin3x\left(sinx-2\right)=0\)

<=> \(\left[{}\begin{matrix}sin3x=0\\sinx-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=k\pi,k\in Z\\sinx=2\end{matrix}\right.\)

Do sin\(\in\left[-1,1\right]\)nên sinx =2 ( loại)

\(3x=k\pi\Leftrightarrow x=k\dfrac{\pi}{3},k\in Z\)

\(cos3x.cosx-cos4x=2-4sin^2\left(\dfrac{\pi}{4}-\dfrac{3x}{2}\right)\)

\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x-cos4x=2cos\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x=2sin3x\)

\(\Leftrightarrow sin3x.sinx=2sin3x\)

\(\Leftrightarrow sin3x.\left(sinx-2\right)=0\)

\(\Leftrightarrow sin3x=0\)

\(\Leftrightarrow3x=k\pi\)

\(\Leftrightarrow x=\dfrac{k\pi}{3}\)

ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)

\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)

\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)

\(\Leftrightarrow2cos^22x-2cos2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)

Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)

\(\begin{array}{l}a)\;\,cos(x + \frac{\pi }{3}) = \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow cos\left( {x + \frac{\pi }{3}} \right) = cos\frac{\pi }{6}\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{3} = \frac{\pi }{6} + k2\pi \\x + \frac{\pi }{3} = -\frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = -\frac{\pi }{6} + k2\pi \\x = -\frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}b)\;\,cos4x = cos\frac{{5\pi }}{{12}}\\ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{{5\pi }}{{12}} + k2\pi \\4x = -\frac{{5\pi }}{{12}} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{48}} + k\frac{\pi }{2}\\x = -\frac{{5\pi }}{{48}} + k\frac{\pi }{2}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}c)\;\,co{s^2}x = 1\\ \Leftrightarrow \left[ \begin{array}{l}cosx = 1\\cosx = -1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \pi  + k2\pi \end{array} \right. \Leftrightarrow x = k\pi ,k \in \mathbb{Z}\end{array}\)

\(ĐK:x\ge0\\ PT\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}=5\\ \Leftrightarrow x=25\left(tm\right)\)

Lời giải:
ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 3(\sqrt{x}-1)=\sqrt{x}+2$

$\Leftrightarrow 2\sqrt{x}=5$

$\Leftrightarrow \sqrt{x}=\frac{5}{2}$

$\Leftrightarrow x=\frac{25}{4}$ (thỏa mãn)

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)

\(\dfrac{\sqrt{x+1}\left(\sqrt{x+1}+\sqrt{3-x}\right)}{2\left(x-1\right)}>x-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)

- TH1: Với \(x>1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)

Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)

BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)

\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)

TH2: với \(x< 1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)

Tương tự như trên, đặt  \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)

\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)