Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(\Leftrightarrow\cos\left(\frac{\pi}{7}-3x\right)=\cos\left(-\frac{5}{6}\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{\pi}{7}-3x=-\frac{5}{6}\pi+k2\pi\\\frac{\pi}{7}-3x=\frac{5}{6}\pi+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{41}{126}\pi-\frac{2}{3}k\pi\\x=-\frac{29}{42}\pi-\frac{2}{3}k\pi\end{matrix}\right.\)
b/ \(\Leftrightarrow\sin\left(90^0-\frac{x}{3}\right)=\sin\left(2x+30^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}90^0-\frac{x}{3}=2x+30^0+k180^0\\90^0-\frac{x}{3}=180^0-2x-30^0+k180^0\end{matrix}\right.\Leftrightarrow...\)
c/ \(DKXD:\cos\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne90^0+k180^0\Leftrightarrow x\ne-30^0-k90^0\)
\(\Leftrightarrow30^0-2x=60^0+k180^0\Leftrightarrow x=-15^0-k90^0\left(tm\right)\)
d/ \(DKXD:\sin\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne k180^0\Leftrightarrow x\ne15^0-k90^0\)
\(\Leftrightarrow30^0-2x=30^0+k.180^0\Leftrightarrow x=-k.90^0\left(tm\right)\)
\(sina+sinb+sinc+3=0\)
\(\Leftrightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)=0\)
Do \(\left\{{}\begin{matrix}sina\ge-1\\sinb\ge-1\\sinc\ge-1\end{matrix}\right.\) ;\(\forall a;b;c\)
\(\Rightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)\ge0\)
Dấu "=" xảy ra khi và chỉ khi \(sina=sinb=sinc=-1\)
\(\Rightarrow cosa=cosb=cosc=0\Rightarrow cosa+cosb+cosc+10=10\)
b/ \(sinx=1-sin^2x\Rightarrow sinx=cos^2x\)
\(\Rightarrow sin^2x=cos^4x\Rightarrow1-cos^2x=cos^4x\)
\(\Rightarrow cos^4x+cos^2x=1\Rightarrow\left(cos^4x+cos^2x\right)^2=1\)
\(\Rightarrow cos^8x+2cos^6x+cos^4x=1\)
Do \(\alpha\in\left(\frac{\pi}{2};\frac{3\pi}{4}\right)\Rightarrow sin\alpha>0;cos\alpha< 0;tan\alpha< 0\)
\(\frac{tana}{cota}=\frac{\sqrt{5}-1}{\sqrt{5}+1}\Leftrightarrow tan^2a=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{\left(\sqrt{5}-1\right)^2}{4}\Rightarrow tana=\frac{1-\sqrt{5}}{2}\Rightarrow cota=\frac{-1-\sqrt{5}}{2}\)
\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{5+\sqrt{5}}{10}\)
\(\Rightarrow sin^2a=1-cos^2a=\frac{5-\sqrt{5}}{10}\)
\(sin2a=2sina.cosa=2tana.cos^2a=-\frac{2\sqrt{5}}{5}\)
Thay vào ta được:
\(P=...\)
Bạn tự thay số và bấm máy
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.
\(cos3x+cosx+cos2x=0\)
\(\Leftrightarrow2cos2x.cosx+cos2x=0\)
\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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; ...
\(\Leftrightarrow tanx+tan2x=1-tanx.tan2x\)
\(\Leftrightarrow\frac{tanx+tan2x}{1-tanx.tan2x}=1\)
\(\Leftrightarrow tan3x=1\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)
Nhân 2 vế với \(sin4x\) sau đó tách:
\(\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=\frac{2sin2x.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}=\frac{4sinx.cosx.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}\)
Rồi rút gọn
a/ Hàm xác định trên R
\(y\left(-x\right)=sin^2\left(-2x\right)+1=sin^22x+1=y\left(x\right)\)
Hàm chẵn
b/ Hàm xác định trên R
\(y\left(-x\right)=sin^2\left(-x\right)-cos^2\left(-x\right)=sin^2x-cos^2x=y\left(x\right)\)
Hàm chẵn
c/ Hàm xác định trên R
\(y=sin^2x+cos^2x=1\Rightarrow y\left(-x\right)=1=y\left(x\right)\)
Hàm chẵn
d/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
Miền xác định của hàm là miền đối xứng
\(y\left(-x\right)=tan\left(-x\right)+3sin\left(-x\right)-7\)
\(=-tanx-3sinx-7\)
Hàm ko chẵn ko lẻ
\(A=\dfrac{tan^2a-sin^2a}{cot^2a-cos^2a}\)
\(A=\dfrac{\dfrac{sin^2a}{cos^2a}-sin^2a}{\dfrac{cos^2a}{sin^2a}-cos^2a}=\dfrac{sin^2a\left(\dfrac{1}{cos^2a}-1\right)}{cos^2a\left(\dfrac{1}{sin^2a}-1\right)}\)
\(A=\dfrac{sin^2a\left(\dfrac{1-cos^2a}{cos^2a}\right)}{cos^2a\left(\dfrac{1-sin^2a}{sin^2a}\right)}=\dfrac{sin^2a\left(\dfrac{sin^2a}{cos^2a}\right)}{cos^2a\left(\dfrac{cos^2a}{sin^2a}\right)}\)
\(A=\dfrac{\dfrac{sin^4a}{cos^2a}}{\dfrac{cos^4a}{sin^2a}}=\dfrac{sin^4a}{cos^2a}.\dfrac{sin^2a}{cos^4a}\)
\(A=\dfrac{sin^6a}{cos^6a}=tan^6a\)