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NV
23 tháng 9 2019

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

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

Đặt \(tan^2x=a>0\)

\(a-\frac{2}{a}+2=0\)

\(\Leftrightarrow a^2+2a-2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=\sqrt{3}-1\\a=-\sqrt{3}-1< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow tan^2x=\sqrt{3}-1\Rightarrow tanx=\pm\sqrt{\sqrt{3}-1}=tan\left(\pm\alpha\right)\)

\(\Rightarrow x=\pm\alpha+k\pi\)

23 tháng 9 2019

Cho em hỏi sao lại có +2 ạ.

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

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

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

19 tháng 9 2021

b)

(sin2x + cos2x)cosx + 2cos2x - sinx = 0

⇔ cos2x (cosx + 2) + sinx (2cos2 x – 1) = 0

⇔ cos2x (cosx + 2) + sinx.cos2x = 0

⇔ cos2x (cosx + sinx + 2) = 0

⇔ cos2x  = 0

⇔ 2x =  + kπ ⇔ x =  + k  (k ∈ )

19 tháng 9 2021

c) 

Đáp án:

x=π6π6+ k2ππ

và x= 5π65π6+k2ππ (k∈Z)

Lời giải:

sin2x-cos2x+3sinx-cosx-1=0

⇔ 2sinxcosx-(1-2sin²x) +3sinx-cosx-1=0

⇔ 2sin²x+2sinxcosx+3sinx-cosx-2=0

⇔ (2sin²x+3sinx-2)+ cosx(2sinx-1)=0

⇔ (2sinx-1)(sinx+2)+cosx(2sinx-1)=0

⇔ (2sinx-1)(sinx+cosx+2)=0

⇔ sinx=1212

⇔ x=π6π6+ k2ππ

hoặc x= 5π65π6+k2ππ (k∈Z)

(sinx+cosx+2)=0 (vô nghiệm do sinx+cosx+2=√22sin(x+π4π4)+2>0)

a: tan x(cot^2x-1)

\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)

=cotx-tanx/cotx=cotx(1-tan^2x)

b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)

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

c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)

\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)

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

=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)

=-cos^2x*cos^2x=-cos^4x

=>ĐPCM

NV
4 tháng 10 2020

1.

\(\Leftrightarrow3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)

2.

\(\Leftrightarrow cos5x=0\Leftrightarrow5x=\frac{\pi}{2}+k\pi\Leftrightarrow x=\frac{\pi}{10}+\frac{k\pi}{5}\)

4.

\(cos3x+cosx+cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos2x=0\)

\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

NV
4 tháng 10 2020

5.

\(sin6x+sin2x+sin4x=0\)

\(\Leftrightarrow2sin4x.cos2x+sin4x=0\)

\(\Leftrightarrow sin4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\)

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

6. ĐKXĐ; ...

\(\Leftrightarrow tanx+tan2x=1-tanx.tan2x\)

\(\Leftrightarrow\frac{tanx+tan2x}{1-tanx.tan2x}=1\)

\(\Leftrightarrow tan3x=1\)

\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)

17 tháng 8 2021

ĐK: \(x\ne k\pi\)

\(\dfrac{1+sin2x+cos2x}{1+cot^2x}=sinx.\left(sin2x+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{cos^2x+sin^2x}{sin^2x}}=sinx.\left(2sinx.cosx+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{1}{sin^2x}}=2sin^2x.\left(cosx+sinx\right)\)

\(\Leftrightarrow1+sin2x+cos2x=2cosx+2sinx\)

\(\Leftrightarrow1+2sinx.cosx+2cos^2x-1=2cosx+2sinx\)

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

\(\Leftrightarrow\left(cosx-1\right).sin\left(x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

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

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

25 tháng 8 2021

Cảm ơn bạn nhé

HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023

a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi  - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)

b) \(2\cos x =  - \sqrt 2 \;\; \Leftrightarrow \cos x =  - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x =  - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)

c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)

\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)

NV
21 tháng 1 2021

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{cosx}{sinx}-1=\dfrac{cos^2x-sin^2x}{1+\dfrac{sinx}{cosx}}+sin^2x-sinx.cosx\)

\(\Leftrightarrow\dfrac{cosx-sinx}{sinx}=cosx\left(cosx-sinx\right)-sinx\left(cosx-sinx\right)\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(\dfrac{1}{sinx}-cosx+sinx\right)=0\)

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

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-sin2x-cos2x\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\right)=0\)