a: cos3x=8

mà -1<=cos3x<=1

nên \(x\in\varnothing\)

b; \(-2\cdot cosx+\sqrt{3}=0\)

=>\(-2\cdot cosx=-\sqrt{3}\)

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

=>x=pi/6+k2pi hoặc x=-pi/6+k2pi

c: cos(3x-pi/6)=0

=>3x-pi/6=pi/2+k2pi

=>3x=2/3pi+k2pi

=>x=2/9pi+k2pi/3

d: cos(x+2/3pi)=cos(pi/5)

=>x+2/3pi=pi/5+k2pi hoặc x+2/3pi=-pi/5+k2pi

=>x=-7/15pi+k2pi hoặc x=-13/15pi+k2pi

e: cos^2(3x)=4

=>cos3x=2(loại) hoặc cos3x=-2(loại)

Đk:\(cosx\ne\dfrac{1}{2}\) \(\Rightarrow cosx\ne\pm\dfrac{\pi}{3}+k2\pi\);\(k\in Z\)

Pt \(\Leftrightarrow\dfrac{\left(2-\sqrt{3}\right)cosx-\left[1-cos\left(x-\dfrac{\pi}{2}\right)\right]}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx-1+cos\left(\dfrac{\pi}{2}-x\right)=2cosx-1\)

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

\(\Leftrightarrow2sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\) (\(k\in Z\)) kết hợp với đk \(\Rightarrow x=\dfrac{2\pi}{3}+k2\pi\)(\(k\in Z\))

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow x\ne\pm\dfrac{\pi}{3}+k2\pi\)

\(\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)-1=2cosx-1\)

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

\(\Leftrightarrow tanx=\sqrt{3}\)

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

Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{2\pi}{3}+k2\pi\)

d: cos^2x=1

=>sin^2x=0

=>sin x=0

=>x=kpi

a: =>sin 4x=cos(x+pi/6)

=>sin 4x=sin(pi/2-x-pi/6)

=>sin 4x=sin(pi/3-x)

=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi

=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3

b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi

=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi

c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi

=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm

a: \(sin\left(x-\dfrac{\Omega}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

=>\(sin\left(x-\dfrac{\Omega}{4}\right)=sin\left(-\dfrac{\Omega}{4}\right)\)

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

b: \(cos\left(x+\dfrac{\Omega}{4}\right)=cos\left(\dfrac{3}{4}\Omega\right)\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{4}=\dfrac{3}{4}\Omega+k2\Omega\\x+\dfrac{\Omega}{4}=-\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)

c: ĐKXĐ: \(\left\{{}\begin{matrix}2x< >\dfrac{\Omega}{2}+k\Omega\\x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\\x< >\dfrac{1}{6}\Omega+k\Omega\end{matrix}\right.\)

\(tan2x=tan\left(x+\dfrac{\Omega}{3}\right)\)

=>\(2x=x+\dfrac{\Omega}{3}+k\Omega\)

=>\(x=\dfrac{\Omega}{3}+k\Omega\)

d: ĐKXĐ: \(2x< >k\Omega\)

=>\(x< >\dfrac{k\Omega}{2}\)

\(cot2x=-\dfrac{\sqrt{3}}{3}\)

=>\(cot2x=cot\left(-\dfrac{\Omega}{3}\right)\)

=>\(2x=-\dfrac{\Omega}{3}+k\Omega\)

=>\(x=-\dfrac{\Omega}{6}+\dfrac{k\Omega}{2}\)

a: \(\Leftrightarrow sin\left(\dfrac{x}{3}-\dfrac{pi}{4}\right)=sinx\)

=>x/3-pi/4=x+k2pi hoặc x/3-pi/4=pi-x+k2pi

=>2/3x=-pi/4+k2pi hoặc 4/3x=5/4pi+k2pi

=>x=-3/8pi+k3pi hoặc x=15/16pi+k*3/2pi

b: =>(sin3x-sin2x)(sin3x+sin2x)=0

=>sin3x-sin2x=0 hoặc sin 3x+sin 2x=0

=>sin 3x=sin 2x hoặc sin 3x=sin(-2x)

=>3x=2x+k2pi hoặc 3x=pi-2x+k2pi hoặc 3x=-2x+k2pi hoặc 3x=pi+2x+k2pi

=>x=k2pi hoặc x=pi/5+k2pi/5 hoặc x=k2pi/5 hoặc x=pi+k2pi

B