ĐKXĐ: ...

\(\Leftrightarrow cos^2x\left(\frac{sin^2x}{cos^2x}+\frac{sinx}{cosx}\right)=\frac{1}{2}\left(sinx+cosx\right)\)

\(\Leftrightarrow2sinx\left(sinx+cosx\right)-\left(sinx+cosx\right)=0\)

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

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

\(\Leftrightarrow...\)

Câu 2 bạn coi lại đề

3.

\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)

\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)

\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)

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

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

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

\(\Leftrightarrow...\)

4.

Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm

5.

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)

\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)

Xét (1):

\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)

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

\(\Leftrightarrow2sin^3x-sinx-1=0\)

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

\(\Leftrightarrow...\)

6.

\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)

\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)

\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

a.

a/ \(\Leftrightarrow tanx.tan\frac{\pi}{9}-1=tan\frac{\pi}{90}\left(tanx+tan\frac{\pi}{9}\right)\)

\(\Leftrightarrow\frac{tanx+tan\frac{\pi}{9}}{1-tanx.tan\frac{\pi}{9}}=-\frac{1}{tan\frac{\pi}{90}}\)

\(\Leftrightarrow tan\left(x+\frac{\pi}{9}\right)=tan\left(\frac{23\pi}{45}\right)\)

\(\Rightarrow x+\frac{\pi}{9}=\frac{23\pi}{45}+k\pi\)

\(\Rightarrow x=\frac{2\pi}{5}+k\pi\)

Do \(-2\pi< x< 2\pi\Rightarrow-2\pi< \frac{2\pi}{5}+k\pi< 2\pi\)

\(\Rightarrow k=\left\{-2;-1;0;1;2\right\}\)

\(\Rightarrow x=\left\{-\frac{8\pi}{5};-\frac{3\pi}{5};\frac{2\pi}{5};\frac{7\pi}{5};\frac{12\pi}{5}\right\}\)

b/

ĐKXĐ: \(cos2x\ne0\)

\(\Leftrightarrow tan^22x+1+tan^22x=7\)

\(\Leftrightarrow tan^22x=3\)

\(\Rightarrow\left[{}\begin{matrix}tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tan2x=tan60^0\\tan2x=tan\left(-60^0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=60^0+k180^0\\2x=-60^0+k180^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-30^0+k180^0\end{matrix}\right.\)

Bạn tự tìm nghiệm thuộc khoảng đã cho nhé

c/ ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow tan^3x+1+tan^2x+4\sqrt{3}\left(1+tanx\right)=8+7tanx\)

\(\Leftrightarrow tan^2x\left(1+tanx\right)+\left(4\sqrt{3}-7\right)\left(1+tanx\right)=0\)

\(\Leftrightarrow\left(tan^2x-7+4\sqrt{3}\right)\left(1+tanx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tan^2x=7-4\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=tan\left(-\frac{\pi}{4}\right)\\tanx=tan\left(\frac{\pi}{12}\right)\\tanx=tan\left(-\frac{\pi}{12}\right)\end{matrix}\right.\)

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

Bạn tự tìm x thuộc khoảng đã cho

a/

\(\Leftrightarrow cos^3x-sin^3x=cosx+sinx\)

- Với \(cosx=0\Rightarrow sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\) là 1 nghiệm

- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow1-tan^3x=\frac{1}{cos^2x}+tanx.\frac{1}{cos^2x}\)

\(\Leftrightarrow1-tan^3x=1+tan^2x+tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+tan^2x+tanx=0\)

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

\(\Leftrightarrow tanx=0\Rightarrow x=k\pi\)

b/

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}=1+2sinx\)

\(\Leftrightarrow\frac{cosx-sinx}{cosx+sinx}=1+2sinx\)

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

\(\Leftrightarrow sinx+sinx.cosx+sin^2x=0\)

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

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

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

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

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

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

c/

ĐKXĐ: ...

Chia 2 vế cho \(cos^2x\) ta được:

\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)

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

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

\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)

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

\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)

\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)

b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)

=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)

d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)

\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)

=\(\frac{1}{cosx.sinx}=VP\)

e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)

c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)

=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)

\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)

Đây nha bạn