\(\Leftrightarrow cos5x=-sin3x\)

\(\Leftrightarrow cos5x=cos\left(\dfrac{\pi}{2}+3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}+3x+k2\pi\\5x=-\dfrac{\pi}{2}-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{16}+\dfrac{k\pi}{4}\end{matrix}\right.\)

Đáp án D

a2: \(2\cdot\sin17x+\sqrt3\cdot cos5x+\sin5x=0\)

=>\(\sin17x+\frac{\sqrt3}{2}\cdot cos5x+\frac12\cdot\sin5x=0\)

=>\(\sin17x+\sin\left(5x+\frac{\pi}{3}\right)=0\)

=>\(\sin17x=-\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(-5x-\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}17x=-5x-\frac{\pi}{3}+k2\pi\\ 17x=\pi+5x+\frac{\pi}{3}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}22x=-\frac{\pi}{3}+k2\pi\\ 12x=\frac43\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{66}\pi+\frac{k\pi}{11}\\ x=\frac19\pi+\frac{k\pi}{6}\end{array}\right.\)

a3: \(cos7x-\sin5x=\sqrt3\left(cos5x-\sin7x\right)\)

=>\(cos7x+\sqrt3\cdot\sin7x=\sqrt3\cdot cos5x+\sin5x\)

=>\(\frac{\sqrt3}{2}\cdot\sin7x+\frac12\cdot cos7x=\frac{\sqrt3}{2}\cdot cos5x+\frac12\cdot\sin5x\)

=>\(\sin\left(7x+\frac{\pi}{6}\right)=\sin\left(5x+\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\ 7x+\frac{\pi}{6}=\pi-5x-\frac{\pi}{3}+k2\pi=-5x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ 12x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{12}+k\pi\\ x=\frac{1}{24}\pi+\frac{k\pi}{6}\end{array}\right.\)

1,Cho tam giác ABC gọi G là trọng tâm.Đường thẳng d không cắt tam giác ABC.Gọi A',B',C',G' lần lượt là hình chiếu của A,B,C,G trên đường thẳng d.Chứng minh rằng GG'=(AA'+BB'+CC')/3 

Bạn giải dùm mình bài này đc ko

sin3x - cos5x = 0

Vậy phương trình có hai họ nghiệm  (k ∈ Z).

nếu tính bình thường thì ra 2cos4x.cos(-x), sao nó lại mất dấu "-" vậy bạn?

\(\Leftrightarrow2cos4x.cosx+2cos^24x-1+1=0\)

\(\Leftrightarrow2cos4x\left(cos4x+cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x+cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=cos\left(\pi-x\right)\end{matrix}\right.\)

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

1.

\(2cos4x-3=0\)

\(\Leftrightarrow cos4x=\dfrac{3}{2}\)

Mà \(cos4x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

2.

\(cos5x+2=0\)

\(\Leftrightarrow cos5x=-2\)

Mà \(cos5x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

3.

\(cos2x+0,7=0\)

\(\Leftrightarrow cos2x=-\dfrac{7}{10}\)

\(\Leftrightarrow2x=\pm arccos\left(-\dfrac{7}{10}\right)+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{arccos\left(-\dfrac{7}{10}\right)}{2}+k\pi\)

4.

\(cos^22x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-\dfrac{1}{2}\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

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

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

e ms lớp 8

ai k minh giai cho