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a.
\(1-sin^2x+1-2sin^2x+sinx+2=0\)
\(\Leftrightarrow-3sin^2x+sinx+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{4}{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
b. ĐKXĐ; ...
\(5tanx-\frac{2}{tanx}-3=0\)
\(\Leftrightarrow5tan^2x-3tanx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{2}{5}\right)+k\pi\end{matrix}\right.\)
e.
Ko rõ vế phải
f.
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
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)
\(2\sin (2x+70^0)+1=0\Leftrightarrow \sin (2x+\frac{7}{18}\pi)=-\frac{1}{2}=\sin (\frac{-\pi}{6})\)
\(\Rightarrow \left[\begin{matrix} 2x+\frac{7}{18}\pi=\frac{-\pi}{6}+2k\pi\\ 2x+\frac{7}{18}\pi=\frac{7}{6}\pi+2k\pi\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=\frac{-5\pi}{18}+k\pi\\ x=\frac{7}{18}\pi+k\pi\end{matrix}\right.\)
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.
d.
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=0\)
\(\Leftrightarrow sin^2x-cos^2x=0\)
\(\Leftrightarrow-cos2x=0\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
e. Đề thiếu
f.
\(\Leftrightarrow sin2x=\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)\)
\(\Leftrightarrow sin2x=cos^2\frac{x}{2}-sin^2\frac{x}{2}\)
\(\Leftrightarrow sin2x=cosx\)
\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
a.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
b.
\(\Leftrightarrow sin2x=1\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
c.
\(\Leftrightarrow2sin2x.cos2x=-1\)
\(\Leftrightarrow sin4x=-1\)
\(\Leftrightarrow4x=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)
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) \(f'(x)=g(x)\)
\(\Leftrightarrow 6\sin ^22x\cos 2x=4\cos 2x-5\sin 4x\)
\(\Leftrightarrow 3\sin ^22x\cos 2x=2\cos 2x-5\sin 2x\cos 2x\)
\(\Leftrightarrow \cos 2x(3\sin ^22x-2+5\sin 2x)=0\)
\(\Leftrightarrow \cos 2x(3\sin 2x-1)(\sin 2x+2)=0\)
\(\Rightarrow \left[\begin{matrix} \cos 2x=0\\ \sin 2x=\frac{1}{3}\\ \sin 2x=-2\end{matrix}\right.\)
Với \(\cos 2x=0\Rightarrow x=\frac{\pm \pi}{4}+k\pi (k\in\mathbb{Z})\)
Với \(\sin 2x=\frac{1}{3}\Rightarrow x=\frac{1}{2}\arcsin \frac{1}{3}+k\pi \) hoặc \(x=\pi -\frac{1}{2}\arcsin \frac{1}{3}+k\pi\)
Với \(\sin 2x=-2\) thì loại vì $\sin 2x\in [-1;1]$
b) \(f'(x)=g(x)\)
\(\Leftrightarrow -x^2\sin x+4x\cos ^2\frac{x}{2}=x-x^2\sin x\)
\(\Leftrightarrow 4x\cos ^2\frac{x}{2}=x\)
\(\Leftrightarrow x(4\cos ^2\frac{x}{2}-1)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ \cos ^2\frac{x}{2}=\frac{1}{4}\rightarrow \cos \frac{x}{2}=\pm \frac{1}{2}\end{matrix}\right.\)
Với \(\cos \frac{x}{2}=\frac{1}{2}\Rightarrow x=\pm \frac{2\pi}{3}+4k\pi \) với $k$ nguyên.
Với \(\cos \frac{x}{2}=\frac{-1}{2}\Rightarrow x=\frac{-4\pi}{3}+4k\pi \) với $k$ nguyên.