\(\sin3x+\sqrt{3}\cos3x=2\sin2x\)

<=>  \(\frac{1}{2}.sin3x+\frac{\sqrt{3}}{2}.cos3x=sin2x\)

<=>  \(sin\left(3x+\frac{\pi}{3}\right)=sin2x\)

<=>  \(\orbr{\begin{cases}3x+\frac{\pi}{3}=2x+k2\pi\\3x+\frac{\pi}{3}=\pi-2x+k2\pi\end{cases}}\)   \(k\inℤ\)

<=>  \(\orbr{\begin{cases}x=-\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{15}+\frac{k2\pi}{5}\end{cases}}\)

Để mình nghĩ

Bạn ghi yêu cầu đề bài ra nhé.

Nowadays, while a lot of students live and study in good physical condition, there are still many students with visual impairments. They are facing a lot of discrimination in life. They may feel that they are treated less fairly than normal students because of their disabilities. In regular schools, they cannot fully paripate in school activities and are often left behind.What can we do to help them overcome their problems? First, people should pay more attention to them and know what they need. They have different learning difficulties! However, that does not mean that they cannot learn. Therefore, schools should provide them with special materials and tools such as Braille or large print books, talking computers or materials with large print. By giving them these things, we can help them learn better.Secondly, there should be more close contacts between students with visual impairment and non-disabled ones. People should create an atmosphere of friendliness and respect during all activities at school. When working together with students with visual impairments, people will fully understand them and may recognize the talents hidden inside them. This is something that all people in our society should do.It is necessary for all people to give students with visual impairments more opportunities in life, and should integrate them into our community because they are part of our society. Better understanding students with visual impairments and creating more contacts between non-disabled people and disabled people will help students with visual impairments overcome their setbacks.

ủa cái zì vậy bn

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mình cx ko bt luôn

A B C D S O I J H

a) Hình chóp đều S.ABCD có O là tâm đáy, suy ra \(SO\perp\left(ABCD\right)\Rightarrow CB\perp SO\)

Hình vuông ABCD có I,J lần lượt là trung điểm BC,AD, suy ra \(CB\perp IJ\)

Vậy \(CB\perp\left(SIJ\right)\)hay \(\left(SBC\right)\perp\left(SIJ\right).\)

b) Ta có: \(OC=\frac{CD}{\sqrt{2}}=a;SC=2a\Rightarrow\frac{OC}{SC}=\frac{1}{2}\)

\(\hept{\begin{cases}SO\perp\left(ABCD\right)\\C\in\left(ABCD\right)\end{cases}}\Rightarrow\left(SC,ABCD\right)=\widehat{SCO}=arc\cos\left(\frac{OC}{SC}\right)=60^0\)(Vì \(\widehat{SCO}< 90^0\))

b) Lấy H thuộc SI sao cho JH vuông góc SI

\(\hept{\begin{cases}AD||BC\\BC\subset\left(SBC\right)\end{cases}}\Rightarrow AD||\left(SBC\right)\)

\(\Rightarrow d\left(AD,SB\right)=d\left(AD,SBC\right)=d\left(J,SBC\right)\)

Ta thấy: SI là giao tuyến của (SIJ) và (SBC), mà \(\hept{\begin{cases}J\in\left(SIJ\right)\\JH\perp SI\end{cases}\left(H\in SI\right)}\)nên \(JH\perp\left(SBC\right)\)

Ta có \(SO=a\sqrt{3},OI=a\frac{\sqrt{2}}{2}\Rightarrow\cos\widehat{OSI}=\frac{SO}{\sqrt{SO^2+OI^2}}=\frac{\sqrt{42}}{7}\)

Suy ra \(d\left(J,SBC\right)=JH=IJ.\cos\widehat{HJI}=IJ.\cos\widehat{OSI}=\frac{\sqrt{42}a}{7}\)

Vậy \(d\left(AD,SB\right)=\frac{\sqrt{42}a}{7}.\)

Chữa câu c:

\(d\left(AD,SB\right)=JH=IJ.\cos\widehat{HJI}=a\sqrt{2}.\frac{\sqrt{42}}{7}=\frac{2\sqrt{21}a}{7}\)

Có \(cos\left(3x+\frac{\pi}{3}\right)\le1,sin\left(\frac{5\pi}{6}+3x\right)\le1\)

do đó \(cos\left(3x+\frac{\pi}{3}\right)+sin\left(3x+\frac{5\pi}{6}\right)=2\)

\(\Leftrightarrow\hept{\begin{cases}cos\left(3x+\frac{\pi}{3}\right)=1\\sin\left(\frac{5\pi}{6}+3x\right)=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3x+\frac{\pi}{3}=k2\pi,\left(k\inℤ\right)\\\frac{5\pi}{6}+3x=\frac{\pi}{2}+l2\pi,\left(l\inℤ\right)\end{cases}}\)

\(\Leftrightarrow x=\frac{-\pi}{9}+\frac{k2\pi}{3},\left(k\inℤ\right)\)

A B C D S O M N P Q

Lấy P thuộc SC sao cho \(AP\perp SC\)

Ta có: \(\hept{\begin{cases}BC\perp AB\\BC\perp AS\end{cases}\Rightarrow BC\perp\left(SAB\right)\Rightarrow AM\perp BC}\)

\(\hept{\begin{cases}AM\perp BC\\AM\perp SB\end{cases}}\Rightarrow AM\perp\left(SBC\right)\Rightarrow SC\perp AM\)

Tương tự \(SC\perp AN\). Do đó \(SC\perp MAN\)

Vì \(AP\perp SC\)nên P nằm trong mặt phẳng (AMN) hay \(CS\perp\left(AMN\right)\)tại P

Lại có: CA cắt (AMN) tại A, O là trung điểm của CA. Suy ra:

\(d\left(O;AMN\right)=\frac{1}{2}d\left(C,AMN\right)=\frac{CP}{2}=\frac{CA^2}{2CS}=\frac{\left(a\sqrt{2}\right)^2}{2\sqrt{\left(2a\right)^2+\left(a\sqrt{2}\right)^2}}=\frac{\sqrt{6}a}{6}\)

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