a.

\(\Leftrightarrow4cos^32x-3cos2x+3cos2x=\sqrt{2}\)

\(\Leftrightarrow cos^32x=\dfrac{\sqrt{2}}{4}\)

\(\Leftrightarrow cos2x=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{4}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{8}+k\pi\) (\(k\in Z\))

c.

ĐKXĐ: \(\left\{{}\begin{matrix}cos3x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(tan3x.tanx=1\)

\(\Rightarrow tan3x=\dfrac{1}{tanx}\)

\(\Rightarrow tan3x=cotx\)

\(\Rightarrow tan3x=tan\left(\dfrac{\pi}{2}-x\right)\)

\(\Rightarrow3x=\dfrac{\pi}{2}-x+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

b.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x-\left(\dfrac{1}{2}+\dfrac{1}{2}cos6x\right)=0\)

\(\Leftrightarrow cos6x=-cos2x\)

\(\Leftrightarrow cos6x=cos\left(\pi-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=\pi-2x+k2\pi\\6x=2x-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=\pi+k2\pi\\4x=-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

(\(k\in Z\))

ĐKXĐ: \(\left\{{}\begin{matrix}cos3x\ne0\\cosx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\frac{sin3x.sinx}{cos3x.cosx}=1\)

\(\Leftrightarrow sin3x.sinx-cos3x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow2sinx.cosx=0\)

\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)

a: ĐKXĐ: sin 2x<>1

=>2x<>pi/2+k2pi

=>x<>pi/4+kpi

\(\dfrac{cos2x}{sin2x-1}=0\)

=>cos2x=0

=>2x=pi/2+kpi

=>x=pi/4+kpi/2

Kết hợp ĐKXĐ, ta được:

x=3/4pi+k2pi hoặc x=7/4pi+k2pi

b: cos(sinx)=1

=>sin x=kpi

=>sin x=0

=>x=kpi

c: \(2\cdot sin^2x-1+cos3x=0\)

=>cos3x+cos2x=0

=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)

=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)

=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi

=>x=-pi+k2pi hoặc x=pi/5+k2pi/5

e: cos3x=-cos7x

=>cos3x=cos(pi-7x)

=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi

=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2

Nhận thấy tanx=0 không phải nghiệm

\(\Leftrightarrow tan3x=\frac{1}{tanx}\)

\(\Leftrightarrow tan3x=cotx=tan\left(\frac{\pi}{2}-x\right)\)

\(\Rightarrow3x=\frac{\pi}{2}-x+k\pi\)

\(\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)

\(\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\) với \(k=\left\{0;1;2...;7\right\}\)

Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

Đề lỗi!

a: Khi m=2 thì pt sẽ là x^2-6x-3=0

=>\(x=3\pm2\sqrt{3}\)

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

4x2 – 1 = (2x + 1)(3x – 5)

⇔ 4x2 – 1 – (2x + 1)(3x – 5) = 0

⇔ (2x – 1)(2x + 1) – (2x + 1)(3x – 5) = 0

⇔ (2x + 1)[(2x – 1) – (3x – 5)] = 0

⇔ (2x + 1)(2x – 1 – 3x + 5) = 0

⇔ (2x + 1)(4 – x) = 0

⇔ 2x + 1= 0 hoặc 4 – x = 0

   + 2x + 1 = 0 ⇔ 2x = -1 ⇔ x = -1/2.

   + 4 – x = 0 ⇔ x = 4.

Vậy phương trình có tập nghiệm