1.

\(\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)=3-4\left(1-sin^2x\right)\)

\(\Leftrightarrow\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)=4sin^2x-1\)

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

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

\(\Leftrightarrow2cos2x\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

2.

ĐKXĐ: ...

\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)

\(\Leftrightarrow\frac{\pi}{4}-x=-\frac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)

3.

\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)

\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)

\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

4.

\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)

\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)

\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)

Đến đây là dễ r nha bn :3

a. ĐKXĐ: ...

Ta có: \(\left\{{}\begin{matrix}VT=\left(tanx-cotx\right)^2+2\ge2\\VP=1+cos^2\left(3x+\frac{\pi}{4}\right)\le2\end{matrix}\right.\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}tanx-cotx=0\\cos^2\left(3x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cos2x=0\\sin\left(3x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)

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

b.

\(\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi\)

\(\Leftrightarrow sinx-1=3k\)

\(\Leftrightarrow sinx=3k+1\)

Do \(-1\le sinx\le1\)

\(\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0\)

\(\Rightarrow k=0\)

\(\Rightarrow sinx=1\)

\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)

c.

ĐKXĐ: ...

\(\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi\)

\(\Leftrightarrow cosx-1=4k-1\)

\(\Leftrightarrow cosx=4k\)

Mà \(-1\le cosx\le1\Rightarrow-1\le4k\le1\)

\(\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0\)

\(\Rightarrow cosx=0\)

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

1/ \(cosx=\frac{1}{3}\Rightarrow x=\pm a+k2\pi\) với \(cosa=\frac{1}{3}\)

Tổng các nghiệm:

\(\sum x=a+a+2\pi+\left(-a+2\pi\right)+\left(-a+4\pi\right)=8\pi\)

2/ ĐKXĐ:...

\(\Leftrightarrow1+tan^2x-2tanx-4=0\)

\(\Leftrightarrow tan^2x-2tanx-3=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arctan3+k\pi\end{matrix}\right.\)

b/ Không hiểu đề đoạn này \(sinx.cosx\left(x+\frac{\pi}{2}\right)\) , góc trong ngoặc không biết là của cái gì?

c/ ĐKXĐ:...

\(1+cot^2x+3tan^2x=5\)

\(\Leftrightarrow\frac{1}{tan^2x}+3tan^2x-4=0\)

\(\Leftrightarrow3tan^4x-4tan^2x+1=0\)

\(\Rightarrow\left[{}\begin{matrix}tan^2x=1\\tan^2x=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}tanx=\pm1\\tanx=\pm\frac{1}{\sqrt{3}}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{4}+k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

d/

ĐKXĐ: \(sinx\ne0\Rightarrow cosx\ne\pm1\)

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

\(\Leftrightarrow2cos^2x+cosx-1=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=-1\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow cosx=cos\frac{\pi}{3}\)

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

a/ ĐKXĐ: \(cosx\ne-\frac{1}{2}\)

\(\Leftrightarrow2cosx-1=6cosx+3\)

\(\Leftrightarrow4cosx=-4\Rightarrow cosx=-1\)

\(\Rightarrow x=\pi+k2\pi\)

b/

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

\(\Leftrightarrow cosx\left(2cos2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=2\left(vn\right)\end{matrix}\right.\)

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

c/

\(\Leftrightarrow2sin2x.cos2x=0\)

\(\Leftrightarrow sin4x=0\)

\(\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}\)

a/

\(\Leftrightarrow2cos^2x-1+cosx+1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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

b/ ĐKXĐ: ...

\(\Leftrightarrow tanx+\frac{1}{tanx}=2\)

\(\Leftrightarrow tan^2x+1=2tanx\)

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

\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)

c/

\(a+b+c=1+\sqrt{3}-1-\sqrt{3}=0\)

\(\Rightarrow\) Pt có 2 nghiệm: \(\left[{}\begin{matrix}tanx=1\\tanx=-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

d/ ĐKXĐ: ...

\(\Leftrightarrow cot^22x+3.cot2x+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cot2x=-1\\cot2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{4}+k\pi\\2x=arccot\left(-2\right)+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{1}{2}arccot\left(-2\right)+\frac{k\pi}{2}\end{matrix}\right.\)