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NV
6 tháng 7 2021

1.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)

\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)

\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)

2.

Đề bài thiếu, cos?x

Và x thuộc khoảng nào?

3.

\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)

\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)

\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)

4.

\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)

\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)

1B

2A

3A

4C

14 tháng 8 2017

a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)

\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)

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

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)

14 tháng 8 2017

b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

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

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

\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx+\sqrt{3}\cdot cosx=0\)

\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)

\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)

(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)

\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)

\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)

4 tháng 11 2023

\(sin(\dfrac{\pi}{2}-x)cot(\pi+x)=cosxcotx=\dfrac{cosx}{tanx}\\ =\dfrac{\dfrac{1}{\sqrt5}}{-2}=\dfrac{-\sqrt5}{10}\)

AH
Akai Haruma
Giáo viên
18 tháng 8 2021

Lời giải:

$-\frac{4}{5}=\cos 2x=2\cos ^2x-1$

$\Leftrightarrow \cos ^2x=\frac{1}{10}$

Vì $x\in (\frac{\pi}{4}; \frac{\pi}{2})$ nên $\cos x>0$

$\Rightarrow \cos x=\sqrt{\frac{1}{10}}$

$\sin^2x=1-\cos ^2x=\frac{9}{10}$
Vì $x\in (\frac{\pi}{4}; \frac{\pi}{2})$ nên $\sin x>0$

$\Rightarrow \sin x=\frac{3}{\sqrt{10}}$

$\sin (x+\frac{\pi}{3})=\sin x\cos \frac{\pi}{3}+\cos x\sin \frac{\pi}{3}$

$=\sqrt{\frac{9}{10}}.\frac{1}{2}+\sqrt{\frac{1}{10}}.\frac{\sqrt{3}}{2}=\frac{\sqrt{30}+3\sqrt{10}}{20}$

19 tháng 8 2021

cam on Akai Haruma

 

a: ĐKXĐ: 2*sin x+1<>0

=>sin x<>-1/2

=>x<>-pi/6+k2pi và x<>7/6pi+k2pi

b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)

mà 1+cosx>=0

nên 2-cosx>=0

=>cosx<=2(luôn đúng)

c ĐKXĐ: tan x>0

=>kpi<x<pi/2+kpi

d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)

=>cos(x-pi/4)<>1/2

=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi

=>x<>7/12pi+k2pi và x<>-pi/12+k2pi

e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi

=>x<>5/6pi+kpi và x<>kpi-pi/4

f: ĐKXĐ: cos^2x-sin^2x<>0

=>cos2x<>0

=>2x<>pi/2+kpi

=>x<>pi/4+kpi/2

 

11 tháng 10 2023

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