Question

1/

\(2sin\frac{x}{2}\left(sin\frac{x}{2}+cos\frac{3x}{2}\right)-cos2x=0\)

2/

\(2cos^2x=1+sin3x+cos4x\)

3/

\(\sqrt{3}cos7x-2sin4xcos3x-sinx=0\)

4/

\(tanx+cotx=2\left(sin2x+cos2x\right)\)

5/

\(sin\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin\left(\frac{\pi}{6}-x\right)=1\)

6/

\(sinx-\left(2-\sqrt{3}\right)cosx=1\)

7/

\(\frac{1}{cosx}+\left(\sqrt{3}-1\right)cosx=\left(\sqrt{3}+1\right)sinx\)

8/

\(sinx-\sqrt{3}cosx=2cos2x\)

9/

\(sin2x+cos2x+3sinx-cosx-2=0\)

10/\(8sin^2x+\sqrt{3}sin2x-6cos^2x+1=0\)

Answer 1

1.

\(\Leftrightarrow2sin^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{3x}{2}-cos2x=0\)

\(\Leftrightarrow1-cosx+sin2x-sinx-cos2x=0\)

\(\Leftrightarrow1-cosx+2sinx.cosx-sinx-1+2sin^2x=0\)

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

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

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

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

Answer 2

2.

\(\Leftrightarrow cos4x-\left(2cos^2x-1\right)+sin3x=0\)

\(\Leftrightarrow cos4x-cos2x+sin3x=0\)

\(\Leftrightarrow-2sin3x.sinx+sin3x=0\)

\(\Leftrightarrow sin3x\left(1-2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sinx=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{3}\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Answer 3

3.

\(\sqrt{3}cos7x-sin7x-sinx-sinx=0\)

\(\Leftrightarrow\sqrt{3}cos7x-sin7x=2sinx\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}cos7x-\frac{1}{2}sin7x=sinx\)

\(\Leftrightarrow cos\left(7x+\frac{\pi}{6}\right)=sinx\)

\(\Leftrightarrow cos\left(7x+\frac{\pi}{6}\right)=cos\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\7x+\frac{\pi}{6}=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+\frac{k\pi}{4}\\x=-\frac{\pi}{9}+\frac{k\pi}{3}\end{matrix}\right.\)

Answer 4

4.

ĐKXĐ: \(sin2x\ne0\)

\(\frac{sinx}{cosx}+\frac{cosx}{sinx}=2\left(sin2x+cos2x\right)\)

\(\Leftrightarrow\frac{sin^2x+cos^2x}{sinx.cosx}=2\left(sin2x+cos2x\right)\)

\(\Leftrightarrow\frac{2}{sin2x}=2\left(sin2x+cos2x\right)\)

\(\Leftrightarrow2sin2x\left(sin2x+cos2x\right)=2\)

\(\Leftrightarrow2sin^22x+2sin2x.cos2x=2\)

\(\Leftrightarrow1-cos4x+sin4x=2\)

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

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

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

\(\Leftrightarrow...\)

Answer 5

5.

Đặt \(x+\frac{\pi}{3}=t\Rightarrow x=t-\frac{\pi}{3}\Rightarrow\frac{\pi}{6}-x=\frac{\pi}{2}-t\)

Pt trở thành:

\(sint+\sqrt{3}sin\left(\frac{\pi}{2}-t\right)=1\)

\(\Leftrightarrow sint+\sqrt{3}cost=1\)

\(\Leftrightarrow\frac{1}{2}sint+\frac{\sqrt{3}}{2}cost=\frac{1}{2}\)

\(\Leftrightarrow sin\left(t+\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}t+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\t+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

Answer 6

6.

\(\Leftrightarrow\frac{1}{\sqrt{8-4\sqrt{3}}}sinx-\frac{2-\sqrt{3}}{\sqrt{8-4\sqrt{3}}}cosx=\frac{1}{\sqrt{8-4\sqrt{3}}}\)

\(\Leftrightarrow sinx.cos\left(\frac{\pi}{12}\right)-cosx.sin\left(\frac{\pi}{12}\right)=cos\left(\frac{\pi}{12}\right)\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{12}\right)=sin\left(\frac{5\pi}{12}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{12}=\frac{5\pi}{12}+k2\pi\\x-\frac{\pi}{12}=\frac{\pi}{12}+k2\pi\end{matrix}\right.\)

7.

ĐKXĐ: ...

Chia 2 vế cho \(cosx\) ta được:

\(\frac{1}{cos^2x}+\sqrt{3}-1=\left(\sqrt{3}+1\right)tanx\)

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

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

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

Answer 7

8.

\(\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=cos2x\)

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

\(\Leftrightarrow cos2x=cos\left(\frac{5\pi}{6}-x\right)\)

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

9.

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

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

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

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

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

Answer 8

10.

\(\Leftrightarrow8sin^2x+2\sqrt{3}sinx.cosx-6cos^2x+1=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)

\(8tan^2x+2\sqrt{3}tanx-6+\frac{1}{cos^2x}=0\)

\(\Leftrightarrow8tan^2x+2\sqrt{3}tanx-6+1+tan^2x=0\)

\(\Leftrightarrow9tan^2x+2\sqrt{3}tanx-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=-\frac{5\sqrt{3}}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=arctan\left(-\frac{5\sqrt{3}}{9}\right)+k\pi\end{matrix}\right.\)