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NV
18 tháng 8 2020

c/

\(\Leftrightarrow1-cos^2\frac{x}{2}-2cos\frac{x}{2}+2=0\)

\(\Leftrightarrow cos^2\frac{x}{2}+2cos\frac{x}{2}-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=1\\cos\frac{x}{2}=-3< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\frac{x}{2}=k2\pi\)

\(\Leftrightarrow x=k4\pi\)

d/ ĐKXĐ: ...

\(\Leftrightarrow tanx-\frac{2}{tanx}+1=0\)

\(\Leftrightarrow tan^2x+tanx-2=0\)

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

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

NV
18 tháng 8 2020

a/

\(\Leftrightarrow\left(cosx-1\right)\left(2cosx-1\right)=0\)

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

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

b \(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\pm\frac{3\pi}{8}+k\pi\end{matrix}\right.\)

NV
31 tháng 7 2020

c/

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow sinx=\frac{\sqrt{2}}{2}\)

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

d/

\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)

\(\Leftrightarrow sin2x+cos2x=1\)

\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

NV
31 tháng 7 2020

a/ ĐKXĐ:...

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

\(\Leftrightarrow sinx-\sqrt{2}=cosx\)

\(\Leftrightarrow sinx-cosx=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{3\pi}{4}+k2\pi\)

b/

ĐKXĐ: ...

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)

\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)

\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)

16 tháng 8 2021

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

<=> 2cos2x+(2-4sinx)cosx+7sinx-5=0

- sinx=1 => 2cos2x-2cosx+2=0 

pt trên vn

16 tháng 8 2021

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

19 tháng 9 2017

hộ vs ae ơi

NV
16 tháng 9 2020

c.

\(\Leftrightarrow2sin2x.cos2x+\sqrt{3}sin2x=0\)

\(\Leftrightarrow sin2x\left(2cos2x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{5\pi}{12}+k\pi\\x=-\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

d.

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-\sqrt{2}< -1\left(l\right)\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

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

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

NV
16 tháng 9 2020

a.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{5}{\sqrt{3}}>1\left(l\right)\end{matrix}\right.\)

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

b.

\(\Leftrightarrow\frac{1}{2}sin4x.cos4x+\frac{1}{8}=0\)

\(\Leftrightarrow\frac{1}{4}sin8x+\frac{1}{8}=0\)

\(\Leftrightarrow sin8x=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{6}+k2\pi\\8x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{48}+\frac{k\pi}{4}\\x=\frac{7\pi}{48}+\frac{k\pi}{4}\end{matrix}\right.\)

NV
10 tháng 4 2019

1/

\(tanx=\frac{sinx}{cosx}=\frac{sin^2x}{sinx.cosx}=\frac{2sin^2x}{2sinx.cosx}\)

\(=\frac{2\left(\frac{1-cos2x}{2}\right)}{sin2x}=\frac{1-cos2x}{sin2x}\)

2/

\(\frac{sin\left(60-x\right)cos\left(30-x\right)+cos\left(60-x\right)sin\left(30-x\right)}{sin4x}=\frac{sin\left(60-x+30-x\right)}{sin4x}=\frac{sin\left(90-2x\right)}{2sin2x.cos2x}\)

\(=\frac{cos2x}{2sin2x.cos2x}=\frac{1}{2sin2x}\)

3/

\(4cos\left(60+a\right)cos\left(60-a\right)+2sin^2a\)

\(=2\left(cos\left(60+a+60-a\right)+cos\left(60+a-60+a\right)\right)+2sin^2a\)

\(=2cos120+2cos2a+2\left(\frac{1-cos2a}{2}\right)\)

\(=-1+2cos2a+1-cos2a=cos2a\)

NV
15 tháng 10 2020

1.

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

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

\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)

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

Xét (1):

Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

NV
15 tháng 10 2020

2.

\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)

\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)

Xét (1):

Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm

NV
18 tháng 10 2020

e.

\(3\left(1-sin^2x\right)-5sinx-1=0\)

\(\Leftrightarrow-3sin^2x-5sinx+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{3}\\sinx=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)

f.

\(2\left(2cos^2x-1\right)-cosx+7=0\)

\(\Leftrightarrow4cos^2x-cosx+5=0\)

Phương trình vô nghiệm

NV
18 tháng 10 2020

g.

\(\Leftrightarrow\sqrt{2}sin\left(4x+\frac{\pi}{4}\right)=2\)

\(\Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=\sqrt{2}>1\)

Phương trình vô nghiệm

h.

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

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\)

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

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

28 tháng 9 2020

@Nguyễn Việt Lâm giúp em với ạ

NV
29 tháng 9 2020

a/ \(4cos^3x-3cosx-4\left(2cos^2x-1\right)+3cosx-4=0\)

\(\Leftrightarrow4cos^3x-8cos^2x=0\)

\(\Leftrightarrow4cos^2x\left(cosx-2\right)=0\)

\(\Leftrightarrow cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(0< \frac{\pi}{2}+k\pi< 14\Rightarrow-\frac{1}{2}< k< \frac{14-\frac{\pi}{2}}{\pi}\Rightarrow k=\left\{0;1;2;3\right\}\)

\(\Rightarrow x=\left\{\frac{\pi}{2};\frac{3\pi}{2};\frac{5\pi}{2};\frac{7\pi}{2}\right\}\)

b/ Bạn coi lại đề, cái ngoặc thứ 2 thiếu \(\left(2cos\left(???\right)+cosx\right)\)

c/ Bạn coi lại đề, có 2 số hạng \(cos2x\) xuất hiện ở vế trái, cấp 3 chắc ko ai cho kiểu vậy đâu, nếu đúng thế thì người ta cộng luôn thành \(2cos2x\) cho rồi