rút gọn biểu thức:

E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))

\(\cos a\times\sin b=-\dfrac{1}{2}\left[\sin\left(a-b\right)-\sin\left(a+b\right)\right]\)

\(=-\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{-1}{2}\times1=-\dfrac{1}{2}\)

thank you 

a: \(\sin3x+cos2x=1+2\cdot\sin x\cdot cos2x\)

=>sin3x+cos2x=1+sin(x+2x)+sin(x-2x)

=>sin3x+cos2x=1+sin3x-sin x

=>cos2x-1+sin x=0

=>\(1-2\cdot\sin^2x-1+\sin x=0\)

=>\(-2\cdot\sin^2x+\sin x=0\)

=>sin x(2sin x-1)=0

TH1: sin x=0

=>\(x=k\pi\)

TH2: 2sin x-1=0

=>\(\sin x=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)

b: \(\sin^3x+cos^3x=2\cdot\left(\sin^5x+cos^5x\right)\)

=>\(\sin^3x-2\cdot\sin^5x+cos^3x-2\cdot cos^5x=0\)

=>\(\sin^3x\left(1-2\cdot\sin^2x\right)+cos^3x\left(1-2\cdot cos^2x\right)=0\)

=>\(\sin^3x\cdot cos2x-cos^3x\cdot cos2x=0\)

=>\(cos2x\left(\sin^3x-cos^3x\right)=0\)

TH1: cos2x=0

=>\(2x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

TH2: \(\sin^3x-cos^3x=0\)

=>\(\sin^3x=cos^3x\)

=>sin x=cosx

=>\(\sin x-cosx=0\)

=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x-\frac{\pi}{4}\right)=0\)

=>\(x-\frac{\pi}{4}=k\pi\)

=>\(x=\frac{\pi}{4}+k\pi\)

f: ĐKXĐ: \(\begin{cases}\sin x<>0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow x<>\frac{k\pi}{2}\)

\(\frac{\tan x-\sin x}{\sin^3x}=\frac{1}{cosx}\)

=>\(\frac{\frac{\sin x}{cosx}-\sin x}{\sin^3x}=\frac{1}{cosx}\)

=>\(\frac{\frac{1}{cosx}-1}{\sin^2x}=\frac{1}{cosx}\)

=>\(\sin^2x=cosx\cdot\left(\frac{1}{cosx}-1\right)=1-cosx\)

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

=>\(cos^2x-cosx=0\)

=>cosx(cosx-1)=0

TH1: cosx=0

=>\(x=\frac{\pi}{2}+k\pi\) (loại)

TH2: cosx-1=0

=>cosx=1

=>\(x=k2\pi\)

=>sin x=0

=>Loại

a.

\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

.jkilfo,o7m5ijk

 Ta có $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin 5\alpha -2\sin \alpha .\cos 4\alpha -2\sin \alpha .\cos 2\alpha$

$=\sin 5\alpha -\left(\sin 5\alpha -\sin 3\alpha \right)-\left(\sin 3\alpha -\sin \alpha \right)$

$=\sin \alpha .$

Vậy $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin \alpha$

Câu a hạ bậc rồi áp dụng cosa + cosb

Câu b thì mối liên hệ giữa tan với cot là ra

b)

(sin2x + cos2x)cosx + 2cos2x - sinx = 0

⇔ cos2x (cosx + 2) + sinx (2cos2 x – 1) = 0

⇔ cos2x (cosx + 2) + sinx.cos2x = 0

⇔ cos2x (cosx + sinx + 2) = 0

⇔ cos2x  = 0

⇔ 2x =  + kπ ⇔ x =  + k  (k ∈ )

c) 

Đáp án:

x=+ k2

và x= +k2 (k∈Z)

Lời giải:

sin2x-cos2x+3sinx-cosx-1=0

⇔ 2sinxcosx-(1-2sin²x) +3sinx-cosx-1=0

⇔ 2sin²x+2sinxcosx+3sinx-cosx-2=0

⇔ (2sin²x+3sinx-2)+ cosx(2sinx-1)=0

⇔ (2sinx-1)(sinx+2)+cosx(2sinx-1)=0

⇔ (2sinx-1)(sinx+cosx+2)=0

⇔ sinx=

⇔ x=+ k2

hoặc x= +k2 (k∈Z)

(sinx+cosx+2)=0 (vô nghiệm do sinx+cosx+2=sin(x+)+2>0)