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NV
8 tháng 8 2020

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

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

NV
8 tháng 8 2020

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

NV
21 tháng 1 2021

a.

Tổng là cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-sin^2x\end{matrix}\right.\)

Do đó: \(S=\dfrac{u_1}{1-q}=\dfrac{1}{1+sin^2x}\)

b. Tương tự, tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=cos^2x\end{matrix}\right.\)

\(\Rightarrow S=\dfrac{1}{1-cos^2x}=\dfrac{1}{sin^2x}\)

c. Do \(0< x< \dfrac{\pi}{4}\Rightarrow0< tanx< 1\)

Tổng trên vẫn là tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-tanx\end{matrix}\right.\)

\(\Rightarrow S=\dfrac{1}{1+tanx}\)

AH
Akai Haruma
Giáo viên
2 tháng 9 2023

a.

Tìm min:

$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$

Vậy $y_{\min}=2$

----------------

Mặt khác: 

$y=4\sin x(\sin x+1)-8(\sin x+1)+11$

$=(\sin x+1)(4\sin x-8)+11$

$=4(\sin x+1)(\sin x-2)+11$

Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$

$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$

$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$

Vậy $y_{\max}=11$

 

AH
Akai Haruma
Giáo viên
2 tháng 9 2023

b.

$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$

$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$

Vậy $y_{\max}=4$.

---------------------------

Mặt khác:

$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$

$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$

$=(1+\sin x)(3-\sin x)$

Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$

$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$

Vậy $y_{\min}=0$

NV
7 tháng 10 2020

a. ĐKXĐ: ...

\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)

NV
7 tháng 10 2020

b.

\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow4cos^32x-2cos2x-1=0\)

Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề

c. ĐKXĐ: ...

\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)

\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)

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

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

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

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

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

1 tháng 8 2020

\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)

\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)

1 tháng 8 2020

\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)

\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)

\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)

5 tháng 9 2021

1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)

⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)

⇔ 2cos2x - 5cosx + 2 = 0

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

⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên

2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)

⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0 

⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)

⇒ sin4x + cos4x = 48.sin4x . cos4x

⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4

⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4

⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x

⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)

⇔ 1 - 2sin22x = 0

⇔ cos4x = 0

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

 

5 tháng 9 2021

3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)

⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)

⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)

⇔ sin2x - sin22x - (1 + cos4x) = 0

⇔ sin2x - sin22x - 2cos22x = 0

⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0

⇔ sin22x + sin2x - 2 = 0

⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)

⇔ sin2x = 1

⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

4, cos5x + cos2x + 2sin3x . sin2x = 0

⇔ cos5x + cos2x + cosx - cos5x = 0

⇔ cos2x + cosx = 0

⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)

⇔ \(cos\dfrac{3x}{2}=0\)

⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)

⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)

Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)

⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}

Vậy các nghiệm thỏa mãn là các phần tử của tập hợp 

\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)

AH
Akai Haruma
Giáo viên
6 tháng 7 2019

a)

\(4\sin (3x+\frac{\pi}{3})-2=0\Leftrightarrow \sin (3x+\frac{\pi}{3})=\frac{1}{2}=\sin (\frac{\pi}{6})\)

\(\Rightarrow \left[\begin{matrix} 3x+\frac{\pi}{3}=\frac{\pi}{6}+2k\pi \\ 3x+\frac{\pi}{3}=\pi-\frac{\pi}{6}+2k\pi\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-\pi}{18}+\frac{2\pi}{3}\\ x=\frac{\pi}{6}+\frac{2\pi}{3}\end{matrix}\right.\) (k nguyên)

c)

\(\sin (x+\frac{x}{4})-1=0\Leftrightarrow \sin (\frac{5}{4}x)=1=\sin (\frac{\pi}{2})\)

\(\Rightarrow \frac{5}{4}x=\frac{\pi}{2}+2k\pi\Rightarrow x=\frac{2}{5}\pi+\frac{8}{5}k\pi \) (k nguyên)

d)

\(2\sin (2x+70^0)+1=0\Leftrightarrow \sin (2x+\frac{7}{18}\pi)=-\frac{1}{2}=\sin (\frac{-\pi}{6})\)

\(\Rightarrow \left[\begin{matrix} 2x+\frac{7}{18}\pi=\frac{-\pi}{6}+2k\pi\\ 2x+\frac{7}{18}\pi=\frac{7}{6}\pi+2k\pi\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-5\pi}{18}+k\pi\\ x=\frac{7}{18}\pi+k\pi\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
6 tháng 7 2019

f)

\(\cos 2x-\cos 4x=0\)

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

\(\Rightarrow \left[\begin{matrix} x=k\pi\\ x=\frac{k}{3}\pi \end{matrix}\right.\) ( k nguyên)

b,e,g bạn xem lại đề, đơn vị không thống nhất.

TL
1 tháng 12 2019

Chứng minh các biểu thức đã cho không phụ thuộc vào x.

Từ đó suy ra f'(x)=0

a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;

b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;

c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0

d,f(x)=\(\frac{3}{2}\)=>f'(x)=0