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

\(\Leftrightarrow3\sqrt{3}\sin x-4\sqrt{3}\sin^3x-2\sin x.\cos x=0\)

Xét \(\sin x=0\) là nghiệm của pt=> \(x=\pi\)

Xét \(\sin x\ne0\) chia cả 2 vế cho \(\sin x\)

\(\Leftrightarrow3\sqrt{3}-4\sqrt{3}\sin^2x-2\cos x=0\)

\(\Leftrightarrow3\sqrt{3}-4\sqrt{3}+4\sqrt{3}\cos^2x-2\cos x=0\)

\(\Leftrightarrow4\sqrt{3}\cos^2x-2\cos x-\sqrt{3}=0\)

đây là pt b2, bn tự giải và kết luận :3

a.

\(\Leftrightarrow\left[{}\begin{matrix}3x=90^0-x+k360^0\\3x=90^0+x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{45^0}{2}+k90^0\\x=45^0+k180^0\end{matrix}\right.\)

b.

\(\Leftrightarrow cos\left(3x+45^0\right)=cos\left(x-180^0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+45^0=x-180^0+k360^0\\3x+45^0=180^0-x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{225^0}{2}+k180^0\\x=\frac{135^0}{4}+k90^0\end{matrix}\right.\)

c.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

d.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{12}+k\pi\\x=-\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

e.

\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=sin\left(2x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{6}-2x\right)\)

\(\Leftrightarrow2x-\frac{\pi}{4}=\frac{\pi}{6}-2x+k2\pi\)

\(\Leftrightarrow x=\frac{5\pi}{48}+\frac{k\pi}{2}\)

\(\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x+sin\frac{9x}{4}=2\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)+sin\left(\frac{9x}{4}\right)=2\)

Do \(\left\{{}\begin{matrix}sin\left(3x-\frac{\pi}{6}\right)\le1\\sin\left(\frac{9x}{4}\right)\le1\end{matrix}\right.\)

Nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(3x-\frac{\pi}{6}\right)=1\\sin\left(\frac{9x}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2\pi}{9}+\frac{k8\pi}{3}\)

1.

Đề là \(x\in\left(0;\frac{\pi}{4}\right)\) hay \(x\in\left[0;\frac{\pi}{4}\right]\) ?

2.

\(sin3x-4sinx.cos2x=0\)

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

\(\Leftrightarrow2sinx-sin3x=0\)

\(\Leftrightarrow2sinx-3sinx+4sin^3x=0\)

\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)

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

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

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

3.

\(sin^2x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

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

4.

\(\sqrt{3}sin2x+1-cos2x=3\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

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

5.

Ko có 4 đáp án thì làm sao biết, có vô số pt tương đương với pt này :)

6.

\(sinx+cosx-2sinx.cosx+1=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t+1-t^2+1=0\)

\(\Leftrightarrow-t^2+t+2=0\)

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

\(\Rightarrow2sinx.cosx=t^2-1=0\)

\(\Leftrightarrow sin2x=0\)

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

\(\Rightarrow\sqrt{2}.sin\left(3x-\dfrac{\pi}{4}\right)-\sqrt{2}.sin\left(5x-\dfrac{\pi}{3}\right)=0\Leftrightarrow sin\left(3x-\dfrac{\pi}{4}\right)=sin\left(5x-\dfrac{\pi}{3}\right)\)

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

\(\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{19\pi}{96}+\dfrac{k\pi}{4}\end{matrix}\right.\); k\(\in Z\)

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

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

1.

\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)

\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)

\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)

\(\Leftrightarrow4cos^32x=\sqrt{2}\)

\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)

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

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

\(\text{1) }cos^2\left(x-\frac{\pi}{6}\right)-sin^2\left(x-\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+m2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{m2\pi}{3}\\x=\frac{\pi}{6}+n2\pi\end{matrix}\right.\\\Leftrightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3} \)

\(2\text{) }sin^4x-sin^4\left(x+\frac{\pi}{2}\right)=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow sin^2x-cos^2x=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow cos\left(\pi-2x\right)=cos\left(\frac{\pi}{6}-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}\pi-2x=\frac{\pi}{6}-x+m2\pi\\\pi-2x=x-\frac{\pi}{6}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}-m2\pi\\x=\frac{7\pi}{18}-\frac{n2\pi}{3}\end{matrix}\right.\)

\(3\text{) }pt\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+m2\pi\\x=n2\pi\end{matrix}\right.\)

a/

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

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

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

b/

\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)

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

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

\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

c/

\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)

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

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

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

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