a/ Thiếu đề, sau dấu "-" hình như còn gì đó

b/ \(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=sin\left(\frac{\pi}{4}\right)\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)

c/ \(\Rightarrow sin2x=-sinx\Leftrightarrow sin2x=sin\left(-x\right)\)

\(\Rightarrow\left[{}\begin{matrix}2x=-x+k2\pi\\2x=\pi+x+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow sinx.cosx=0\Leftrightarrow sin2x=0\)

\(\Rightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)

e/ f/ Thiếu đề

g/ \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=-cos2x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=cos\left(\pi-2x\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2x+k2\pi\\3x=-2x+k2\pi\\3x=\pi-2x+k2\pi\\3x=2x-\pi+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=-\pi+k2\pi\end{matrix}\right.\)

ĐKXĐ: ...

\(\Leftrightarrow\frac{3cos^2x}{sin^2x}-2cosx+2\sqrt{2}sin^2x-3\sqrt{2}cosx=0\)

\(\Leftrightarrow cosx\left(\frac{3cosx-2sin^2x}{sin^2x}\right)-\sqrt{2}\left(3cosx-2sin^2x\right)=0\)

\(\Leftrightarrow\left(3cosx-2sin^2x\right)\left(\frac{cosx}{sin^2x}-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3cosx-2sin^2x=0\\cosx-\sqrt{2}sin^2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2cos^2x+3cosx-2=0\\\sqrt{2}cos^2x+cosx-\sqrt{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\alpha.\beta=\frac{\pi^2}{12}\)

\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{7}}{4}\)

\(tana=\frac{sina}{cosa}=-\frac{3\sqrt{7}}{7}\) ; \(cota=\frac{1}{tana}=-\frac{\sqrt{7}}{3}\)

\(A=\frac{-\frac{6\sqrt{7}}{7}+\sqrt{7}}{-\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{7}}=\frac{4}{5}\)

a/ \(\pi< a< \frac{3\pi}{2}\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{3}}{2}\)

\(\Rightarrow A=4\left(-\frac{1}{2}\right)^2-2\left(-\frac{\sqrt{3}}{2}\right)+3\left(-\frac{1}{2}\right):\left(-\frac{\sqrt{3}}{2}\right)=1+2\sqrt{3}\)

b/ Bạn viết lại biểu thức, ko biết đâu là tử đâu là mẫu, và góc \(\alpha\) đề có cho nằm ở khoảng nào ko?

ĐK: \(x\ne k\pi\)

Đặt \(\left\{{}\begin{matrix}cotx=a\\sinx=b\end{matrix}\right.\left(a\in R;b\in\left[-1;1\right]\right)\), khi đó:

\(3cot^2x+2\sqrt{2}sin^2x=\left(2+3\sqrt{2}\right)cosx\)

\(\Leftrightarrow3a^2+2\sqrt{2}b^2=\left(2+3\sqrt{2}\right)ab\)

\(\Leftrightarrow3a^2-2ab+2\sqrt{2}b^2-3\sqrt{2}ab=0\)

\(\Leftrightarrow\left(3a-2b\right)\left(a-\sqrt{2}b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3a=2b\\a=\sqrt{2}b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3cotx=2sinx\\cotx=\sqrt{2}sinx\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3cosx=2sin^2x\\cosx=\sqrt{2}sin^2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3cosx=2-2cos^2x\\cosx=\sqrt{2}-\sqrt{2}cos^2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2cos^2x+3cosx-2=0\\\sqrt{2}cos^2x+cosx-\sqrt{2}=0\end{matrix}\right.\)

TH1: \(2cos^2x+3cosx-2=0\Leftrightarrow cosx=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)

TH2: \(\sqrt{2}cos^2x+cosx-\sqrt{2}=0\Leftrightarrow cosx=\dfrac{\sqrt{2}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{4}+k2\pi\)

a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)

\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)

\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)

Đến đây là dễ r nha bn :3

B

a: pi<x<3/2pi

=>sinx<0 và cosx<0

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+\dfrac{9}{4}=\dfrac{13}{4}\)

=>\(cos^2x=\dfrac{4}{13}\)

=>\(\left\{{}\begin{matrix}cosx=-\dfrac{2}{\sqrt{13}}\\sin^2x=\dfrac{9}{13}\end{matrix}\right.\)

mà sin x<0

nên \(sinx=-\dfrac{3}{\sqrt{13}}\)

\(cotx=1:\dfrac{3}{2}=\dfrac{2}{3}\)

b: 0<x<90 độ

=>sin x>0 và cosx>0

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+\dfrac{1}{3}=\dfrac{4}{3}\)

=>\(cos^2x=\dfrac{3}{4}\)

=>\(cosx=\dfrac{\sqrt{3}}{2}\)

=>\(sinx=\dfrac{1}{2}\)

cotx=1:căn 3/3=3/căn 3=căn 3

c: 3/2pi<x<2pi

=>sinx<0 và cosx>0

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+\dfrac{1}{3}=\dfrac{4}{3}\)

=>\(sin^2x=\dfrac{3}{4}\)

mà sin x<0

nên \(sinx=-\dfrac{\sqrt{3}}{2}\)

\(cos^2x=1-\dfrac{3}{4}=\dfrac{1}{4}\)

mà cosx>0

nên cosx=1/2