@Nguyễn Việt Lâm giúp em với ạ
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b.
\(sinx\left(1+2cos^2x-1\right)+2sinx.cosx=1+2cos^2x-1\)
\(\Leftrightarrow cos^2x.sinx+sinx.cosx-cos^2x=0\)
\(\Leftrightarrow cosx\left(sinx.cosx+sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx.cosx+sinx-cosx=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(sinx-cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{1-t^2}{2}+t=0\)
\(\Leftrightarrow-t^2+2t+1=0\Rightarrow\left[{}\begin{matrix}t=1-\sqrt{2}\\t=1+\sqrt{2}>\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1-\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
c/
ĐKXĐ: ....
\(\frac{1}{sinx}+\frac{1}{sin\left(x+\frac{\pi}{2}-2\pi\right)}=4sin\left(2\pi-\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\frac{1}{sinx}+\frac{1}{cosx}=-4sin\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow\frac{sinx+cosx}{sinx.cosx}=-4sin\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow\frac{\sqrt{2}sin\left(x+\frac{\pi}{4}\right)}{sinx.cosx}+4sin\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x=...\\\frac{\sqrt{2}}{sinx.cosx}+4=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}+4sinx.cosx=0\)
\(\Leftrightarrow sin2x=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow x=...\)
d. ĐKXĐ: ...
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)cot\left(\frac{\pi}{2}-\left(x+\frac{\pi}{3}\right)\right)\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)tan\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow1-\frac{1}{4}\left(1-cos4x\right)=\frac{7}{8}\)
\(\Leftrightarrow cos4x=\frac{1}{2}\)
\(\Leftrightarrow x=...\)
