1.

c, \(sin\left(\dfrac{\pi}{3}-x\right)=-\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\\dfrac{\pi}{3}-x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\x=-\dfrac{2\pi}{3}+arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

d, \(sin4x=\dfrac{2}{3}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}arcsin\dfrac{2}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arcsin\dfrac{2}{3}+\dfrac{k\pi}{2}\end{matrix}\right.\)

1.

e, \(2sin2x+\sqrt{2}=0\)

\(\Leftrightarrow sin2x=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin2x=sin\left(-\dfrac{\pi}{4}\right)\)

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

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

1.

f, \(sin\left(3x+\dfrac{\pi}{4}\right)=sin\left(x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=x-\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{4}=-x+\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5\pi}{24}+k\pi\\x=\dfrac{11\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Ta có: ΔABC vuông tại A

nên \(\widehat{B}+\widehat{C}=90^0\)

hay \(\widehat{C}=30^0\)

Xét ΔABC vuông tại A có 

\(AB=BC\cdot\sin30^0\)

\(\Leftrightarrow BC=4:\dfrac{1}{2}=8\left(cm\right)\)

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(AB^2+AC^2=BC^2\)

\(\Leftrightarrow AC^2=8^2-4^2=48\)

hay \(AC=4\sqrt{3}\left(cm\right)\)

Chịuuuuuu~~~~~

Lớp 12 ko bít làm.

a: Xét ΔABC có MN//BC

nên AM/AB=AN/AC

=>AN/4=1,2/3=4/10

hay AN=1,6(cm)

b: BC=5cm

Xét ΔABC có AD là phân giác

nên BD/AB=CD/AC
=>BD/3=CD/4

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{BD}{3}=\dfrac{CD}{4}=\dfrac{BD+CD}{3+4}=\dfrac{5}{7}\)

Do đó: BD=15/7(cm); CD=20/7(cm)

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