(1-4sin^2 x)sin3x=1/2
1+sin(x/2)sinx-cos(x/2)sin^2 x=2cos^2 (pi/4-x/2)
(sin^3 x.sin3x+cos^3 x.cos3x)/tan(x-pi/6)tan(x+pi/3)=-1/8
2cos^2 (pi/4-3x)-4cos4x-15sin2x=21

1 đôi đũa = 2 cái đũa

=2 cái đũa

Chắc vậy :V

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what ?

Ko ghi lại đề nha ! 

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x=-\frac{pi}{6}+k2pi\\2x=pi-\left(-\frac{pi}{6}\right)+k2pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{pi}{12}+kpi\\x=\frac{7}{12}pi+kpi\end{cases}}\)

1.

        \(\cos2x+\sin\left(x+\frac{pi}{4}\right)=0\)

\(\Leftrightarrow\sin\left(x+\frac{pi}{4}\right)=-\cos2x\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{pi}{4}=2x-\frac{pi}{2}+k2pi\\x+\frac{pi}{4}=pi-2x+\frac{pi}{2}+k2pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{3}{4}pi+k2pi\\3x=+\frac{5}{4}pi+k2pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}pi+k2pi\\x=\frac{5}{12}pi+k\frac{2}{3}pi\end{cases}}\)

2.

\(\sin\left(3x-\frac{5pi}{6}\right)+\cos\left(3x+\frac{3pi}{6}\right)=0\)

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

\(\Leftrightarrow\sin\left(3x-\frac{5pi}{6}\right)=\sin\left(3x+\frac{3pi}{6}-\frac{pi}{2}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{5pi}{6}=3x+\frac{3pi}{6}-\frac{pi}{2}+k2pi\\3x-\frac{5pi}{6}=pi-3x-\frac{3pi}{6}+\frac{pi}{2}+k2pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0x=\frac{5pi}{6}+k2pi\left(VN\right)\\6x=\frac{11pi}{6}+k2pi\end{cases}}\)

\(\Leftrightarrow x=\frac{11pi}{36}+k\frac{1}{3}pi\)