Bằng 4 chứ bằng bao nhiêu dễ ẹt ah

tan a=2

=>sin a=2*cosa

\(P=\dfrac{10cosa-3cosa}{cosa+2\cdot2cosa}=\dfrac{7}{5}\)

A = sin⁶x + cos⁶x +sin⁴x +cos⁴x + 5sin²x.cos²x

\(=\left(\sin^2x+\cos^2x\right)\left(\sin^4x-\sin^2x\cos^2x+\cos^4x\right)+\sin^4x+\cos^4x+5\sin^2x\cos^2x\)

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

\(=2\)

Đáp án B

a/

\(\Leftrightarrow cos2x=sin3x\)

\(\Leftrightarrow cos2x=cos\left(\frac{\pi}{2}-3x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k2\pi}{5}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow\left(sinx-1\right)\left(2sinx+1\right)\left(sin^2x-2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{2}\\sinx=1-\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow x=...\)

c/

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1+cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow1-3sin^2x.cos^2x=1+sin2x\)

\(\Leftrightarrow-\frac{3}{4}sin^22x=sin2x\)

\(\Leftrightarrow3sin^22x+4sin2x=0\)

\(\Leftrightarrow sin2x\left(3sin2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=-\frac{4}{3}\left(l\right)\end{matrix}\right.\)

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

a) \(\sqrt{\frac{1+\cos x}{1-\cos x}}-\sqrt{\frac{1-\cos x}{1+\cos x}}=\frac{\sqrt{\left(1+\cos x\right)^2}-\sqrt{\left(1-\cos x\right)^2}}{\sqrt{\left(1-\cos x\right)\left(1+\cos x\right)}}\)

\(=\frac{1+\cos x-1+\cos x}{\sqrt{1-\cos^2x}}=\frac{2\cos x}{\sqrt{\sin^2x}}=\frac{2\cos x}{\sin x}=2\cot x\)

b) \(\frac{1}{\tan x+1}+\frac{1}{\cot x+1}=\frac{\tan x+1+\cot x+1}{\left(\tan x+1\right)\left(\cot x+1\right)}\)

\(=\frac{\tan x+\cot x+2}{\tan x+\cot x+\tan x.\cot x+1}=\frac{\tan x+\cot x+2}{\tan x+\cot x+2}=1\)

c) (ko bt có sai đề ko, làm mãi ko ra) 

d) \(\sin^21^0+\sin^22^0+\sin^23^0+...+\sin^289^0\)

\(=\left(\sin^21^0+\sin^289^0\right)+\left(\sin^22^0+\sin^288^0\right)+...+\sin^245^0\)

\(=\left[\left(\sin^21^0-\cos^289^0\right)+\left(\sin^289^0+\cos^289^0\right)\right]+\)

\(\left[\left(\sin^22^0-\cos^288^0\right)+\left(\sin^288^0+\cos^288^0\right)\right]+...+\sin^245^0\)

\(=\left(0+1\right)+\left(0+1\right)+...+\frac{\sqrt{2}}{2}=\frac{44+\sqrt{2}}{2}\)

MN K BT?

1. 

ĐKXĐ: \(x\ne k\pi\)

\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.

3.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)

\(\Leftrightarrow\dfrac{3\left(1+cosx\right)}{1-cosx}+2\left(1+cosx\right)=0\)

\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{3}{1-cosx}+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

4.

\(\Leftrightarrow\left(sin^2x+cos^2x+2sinx.cosx\right)+\left(sinx+cosx\right)+\left(cos^2x-sin^2x\right)=0\)

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

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

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)