Question

Câu 1: Giải các phương trình sau:

a, \(\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2\)+\(\sqrt{3}cosx=2\)

b, \(\frac{\left(1-2sinx\right).cosx}{\left(1+2sinx\right)\left(1-sinx\right)}=\sqrt{3}\)

c, 5sinx-2=3(1-sinx).tan²x

d, \(\frac{2\left(sin^6x+cos^6\right)}{\sqrt{2}-2sinx}=0\)

e, cos²3x.cos2x-cos²x=0

Câu 2: giải các phương trình sau:

a, sinx+cosx.sin2x+\(\sqrt{3}cos3x=2\left(cos4x+sin^3x\right)\)

b, \(\frac{\left(2-\sqrt{3}\right).cosx-2sin2\left(\frac{x}{2}-\frac{\pi}{4}\right)}{2cosx-1}\)

c, 8sin²2x.cos2x=\(\sqrt{3}sin2x+cos2x\)

d, sin³x- \(\sqrt{3}cos^3x=sinxcos^2x-\sqrt{3}sin^2xcosx\)

Answer 1

a)

Answer 2

B1

b)

Answer 3

@Nguyễn Việt Lâm giúp em với ạ

Answer 4

1.c

ĐKXĐ: ...

\(5sinxx-2=3\left(1-sinx\right).\frac{sin^2x}{cos^2x}\)

\(\Leftrightarrow5sinx-2=\frac{3\left(1-sinx\right).sin^2x}{1-sin^2x}\)

\(\Leftrightarrow5sinx-2=\frac{3\left(1-sinx\right).sin^2x}{\left(1-sinx\right)\left(1+sinx\right)}\)

\(\Leftrightarrow\left(5sinx-2\right)\left(1+sinx\right)=3sin^2x\)

\(\Leftrightarrow2sin^2x+3sinx-2=0\)

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

\(\Leftrightarrow...\)

Answer 5

1d.

Đề ko rõ

1e.

\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)

\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)

\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)

\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)

\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)

\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)

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

Answer 6

2.a

\(sinx+\frac{1}{2}sin3x+\frac{1}{2}sinx+\sqrt{3}cos3x=2cos4x+\frac{3}{2}sinx-\frac{1}{2}sin3x\)

\(\Leftrightarrow sin3x+\sqrt{3}cos3x=2cos4x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}cos3x+\frac{1}{2}sin3x=cos4x\)

\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos4x\)

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

\(\Leftrightarrow...\)

Answer 7

2b.

Đề thiếu

2c.

Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)

\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)

\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)

\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)

\(\Leftrightarrow...\)

Answer 8

Cách khác cho bài 2c:

\(4\left(1-cos4x\right)cos2x=\sqrt{3}sin2x+cos2x\)

\(\Leftrightarrow-4cos4x.cos2x=\sqrt{3}sin2x-3cos2x\)

\(\Leftrightarrow-2cos6x-2cos2x=\sqrt{3}sin2x-3cos2x\)

\(\Leftrightarrow-2cos6x=\sqrt{3}sin2x-cos2x\)

\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=cos6x\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos6x\)

\(\Leftrightarrow...\)

Answer 9

2d.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(tan^3x-\sqrt{3}=tanx-\sqrt{3}tan^2x\)

\(\Leftrightarrow tan^3x+\sqrt{3}tan^2x-tanx-\sqrt{3}=0\)

\(\Leftrightarrow tan^2x\left(tanx+\sqrt{3}\right)-\left(tanx+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(tan^2x-1\right)\left(tanx+\sqrt{3}\right)=0\)

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

\(\Leftrightarrow...\)