Học bài trước rồi à :D

a: A=(sinx+cosx)^2-1=m^2-1

b: B=căn (sinx+cosx)^2-4sinxcosx=căn m^2-4(m^2-1)=căn -3m^2+4

c: C=(sin^2x+cos^2x)^2-2(sinx*cosx)^2=1-2m^2

D) tan²x + cot²x
= (1 - 2)(-sin2x/2 + 1/2)²):(-sin2x/2 + 1/2)²
= (1 - 2sin²x)/sin²x.cos²x
= (m² - 3)/2

cos²x + sin²x=1

=>sin²x=1-cos²x=0.75

=>sinx=\(\pm\)\(\sqrt{3}\)/2

A= \(\dfrac{0,5+2.0,75}{0,5^2\pm\dfrac{\sqrt{3}}{2}}\)= \(\dfrac{-8\pm16\sqrt{3}}{11}\)

\(0< =sin^2x< =1\)

=>\(-2< =sin^2x-2< =-1\)

=>\(sin^2x-2< 0\)

\(0< =cos^2x< =1\)

=>\(-2< =cos^2x-2< =-1\)

\(\Leftrightarrow cos^2x-2< 0\)

\(\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4\cdot sin^2x}\)

\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\cdot\left(1-cos^2x\right)}\)

\(=\sqrt{sin^4x-4sin^xx+4}+\sqrt{cos^4x-4\cdot cos^2x+4}\)

\(=\sqrt{\left(sin^2x-2\right)^2}+\sqrt{\left(cos^2x-2\right)^2}\)

\(=\left|sin^2x-2\right|+\left|cos^2x-2\right|\)

\(=2-sin^2x+2-cos^2x\)

\(=4-\left(sin^2x+cos^2x\right)=4-1=3\)

giúp mk với gấp ạ!!!!

a.Ta có : \(x\in\left(\pi;\dfrac{3}{2}\pi\right)\Rightarrow cosx< 0\) 

\(cosx=-\sqrt{1-sin^2x}=-\sqrt{1-0,8^2}=-0,6\) 

\(tanx=\dfrac{4}{3};cotx=\dfrac{3}{4}\)

b. cos 2x = \(cos^2x-sin^2x=0,6^2-0,8^2=-0,28\)

\(P=2.cos2x=-0,56\)

\(Q=tan\left(2x+\dfrac{\pi}{3}\right)=\dfrac{tan2x+tan\dfrac{\pi}{3}}{1-tan2x.tan\dfrac{\pi}{3}}=\dfrac{tan2x+\sqrt{3}}{1-tan2x.\sqrt{3}}\)

tan 2x = \(\dfrac{2tanx}{1-tan^2x}=\dfrac{\dfrac{2.4}{3}}{1-\left(\dfrac{4}{3}\right)^2}=\dfrac{-24}{7}\) 

\(Q=\dfrac{-\dfrac{24}{7}+\sqrt{3}}{1+\dfrac{24}{7}.\sqrt{3}}\) \(=\dfrac{-24+7\sqrt{3}}{7+24\sqrt{3}}\) 

a)

PT $\Leftrightarrow \sin ^2x-4\sin x\cos x+3\cos ^2x+2(\sin ^2x+\cos ^2x)=2$

$\Leftrightarrow \sin ^2x-4\sin x\cos x+3\cos ^2x=0$

$\Leftrightarrow (\sin x-3\cos x)(\sin x-\cos x)=0$

Nếu $\sin x-3\cos x=0$. Dễ thấy $\sin x, \cos x\neq 0$ nên $\tan x=\frac{\sin x}{\cos x}=3$

$\Rightarrow x=k\pi +\tan ^{-1}(3)$ với $k$ nguyên

Nếu $\sin x=\cos x$ thì tương tự ta có $\tan x=1\Rightarrow x=\pi (k+\frac{1}{4})$ với $k$ nguyên

b)
PT $\Leftrightarrow 25(\sin ^2x+\cos ^2x)+30\sin x\cos x-16\cos ^2x=25$

$\Leftrightarrow 30\sin x\cos x-16\cos ^2x=0$

$\Leftrightarrow \cos x(15\sin x-8\cos x)=0$

Nếu $\cos x=0\Rightarrow x=\pi (k+\frac{1}{2})$ với $k$ nguyên

Nếu $15\sin x-8\cos x=0$

Dễ thấy $\cos x\neq 0$ nên suy ra $\tan x=\frac{\sin x}{\cos x}=\frac{8}{15}$

$\Rightarrow x=k\pi +\tan ^{-1}(\frac{8}{15})$ với $k$ nguyên

c) \(\left\{\begin{matrix} \sin x+\cos x=1\\ \sin ^2x+\cos ^2x=1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} (\sin x+\cos x)^2=1\\ \sin ^2x+\cos ^2x=1\end{matrix}\right.\)

\(\Rightarrow 2\sin x\cos x=0\Leftrightarrow \sin 2x=0\Rightarrow x=\frac{k}{2}\pi\) với $k$ nguyên.