a: pi/2<a<pi

=>sin a>0

\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)

\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)

b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

c: \(sin\left(a-\dfrac{pi}{3}\right)\)

\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)

d: \(cos\left(a-\dfrac{pi}{6}\right)\)

\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)

\(sinx+cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)

\(=\sqrt{2}cos\left(\frac{\pi}{2}-\left(x+\frac{\pi}{4}\right)\right)=\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)

\(sinx-cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx-\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)

\(=-\sqrt{2}sin\left(\frac{\pi}{4}-x\right)=-\sqrt{2}cos\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x\)

\(=sin^2x-cos^2x+sin2x=sin2x-cos2x\)

\(=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)

Bạn ghi ko đúng đề

a/ \(y'=2cos2x=0\Rightarrow cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

Do \(x\in\left[0;\frac{\pi}{2}\right]\Rightarrow x=\frac{\pi}{4}\)

\(cos2x< 0\) khi \(\frac{\pi}{4}< x< \frac{\pi}{2}\); \(cos2x>0\) khi \(0< x< \frac{\pi}{4}\)

Hàm số đồng biến trên \(\left(0;\frac{\pi}{4}\right)\) nghịch biến trên \(\left(\frac{\pi}{4};\frac{\pi}{2}\right)\)

b/ \(y'=-2sin2x=0\Rightarrow sin2x=0\Rightarrow x=\frac{k\pi}{2}\)

Do \(x\in\left(-\frac{\pi}{4};\frac{\pi}{4}\right)\Rightarrow x=0\)

Hàm số đồng biến trên \(\left(-\frac{\pi}{4};0\right)\) nghịch biến trên \(\left(0;\frac{\pi}{4}\right)\)

Bạn ghi đề chính xác ra đi, câu a và câu b đó bạn

Câu a sau \(\frac{4}{cotx}\) còn dấu + nhưng không biết cộng với cái gì

Câu b biểu thức cos đầu tiên là \(cos^2\left(x+\frac{\pi}{3}\right)\) hay \(cos\left(2x+\frac{2\pi}{3}\right)\)

a) Đề thiếu

b)

PT $\Leftrightarrow 1-2\sin^2(x+\frac{\pi}{3})+4\cos (\frac{\pi}{6}-x)-\frac{5}{2}=0$

$\Leftrightarrow 1-2\sin ^2[\frac{\pi}{2}-(\frac{\pi}{6}-x)]+4\cos (\frac{\pi}{6}-x)-\frac{5}{2}=0$

$\Leftrightarrow -2\cos ^2(\frac{\pi}{6}-x)+4\cos (\frac{\pi}{6}-x)-\frac{3}{2}=0$

$\Leftrightarrow -2t^2+4t-\frac{3}{2}=0$ với $t=\cos (\frac{\pi}{6}-x)$

Đến đây bạn giải pt bậc 2 thu được $\cos (\frac{\pi}{6}-x)=\frac{1}{2}$

$\Rightarrow x=2k\pi +\frac{\pi}{2}$ hoặc $x=2k\pi -\frac{\pi}{6}$ với $k$ nguyên

c)

ĐK:.............

PT $\Leftrightarrow 1+\frac{\sin ^2x}{\cos ^2x}-1+\tan x-\sqrt{3}(\tan x+1)=0$

$\Leftrightarrow \tan ^2x+\tan x-\sqrt{3}(\tan x+1)=0$

$\Leftrightarrow \tan ^2x+(1-\sqrt{3})\tan x-\sqrt{3}=0$

$\Rightarrow \tan x=\sqrt{3}$ hoặc $\tan x=-1$

$\Rightarrow x=\pi (k-\frac{1}{4})$ hoặc $x=\pi (k+\frac{1}{3})$ với $k$ nguyên

d)

ĐK:.......

PT $\Leftrightarrow \tan x-\frac{2}{\tan x}+1=0$

$\Leftrightarrow \tan ^2x+\tan x-2=0$

$\Leftrightarrow (\tan x-1)(\tan x+2)=0$

$\Rightarrow \tan x=1$ hoặc $\tan x=-2$

$\Rightarrow x=k\pi +\frac{\pi}{4}$ hoặc $x=k\pi +\tan ^{-2}(-2)$ với $k$ nguyên.

a/ \(\frac{\pi}{6}< x< \frac{\pi}{3}\Rightarrow cosx>0\)

\(cos^2x=\frac{1}{1+tan^2x}=\frac{1}{10}\)

\(cotx=\frac{1}{tanx}=\frac{1}{3}\)

Thay số và bấm máy

b/ \(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\tana< 0\end{matrix}\right.\)

\(sina=\sqrt{1-cos^2a}=\frac{3}{5}\)

\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)

\(A=\frac{6sina.cosa-\frac{2tana}{1-tan^2a}}{cosa-\left(2cos^2a-1\right)}\)

Thay số và bấm máy

c/ \(\frac{3\pi}{2}< x< 2\pi\Rightarrow\left\{{}\begin{matrix}cosx>0\\sinx< 0\end{matrix}\right.\)

\(cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{1}{\sqrt{5}}\)

\(sinx=cosx.tanx=-\frac{2}{\sqrt{5}}\)

\(B=\frac{cos^2x+2sinx.cosx}{\frac{2tanx}{1-tan^2x}-\left(2cos^2x-1\right)}\)

Thay số

\(sin\left(x-\frac{\pi}{2}\right)+sin\frac{13\pi}{2}=sin\left(x+\frac{\pi}{2}\right)\)

\(\Leftrightarrow-cosx+1=cosx\)

\(\Leftrightarrow2cosx=1\Rightarrow cosx=\frac{1}{2}\)

Ta có: \(sin^2x+cos^2x=1\Rightarrow sinx=\sqrt{1-\left(-\frac{2}{3}\right)^2}=\frac{\sqrt{5}}{3}\)

  \(P=\left(1+3sin2x\right)\left(1+4cos2x\right)=\left(1+3.2sinx.cosx\right)\left[1+4.\left(cos^2x-sin^2x\right)\right]\)

     \(=\left(1+6.\frac{\sqrt{5}}{3}.\left(-\frac{2}{3}\right)\right).\left[1+4\left(\left(-\frac{2}{3}\right)^2-\left(\frac{\sqrt{5}}{3}\right)^2\right)\right]\) \(=\frac{15-20\sqrt{5}}{27}\)