Giải phương trình :(cô mik gợi ý đánh giá số mũ)
sin5x + cos5x +sin2x +cos2x =1 +\(\sqrt{2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
15 tháng 8 2020
4.
ĐKXĐ: \(2cos^2x+sinx-1\ne0\)
\(\Leftrightarrow-2sin^2x+sinx+1\ne0\Rightarrow\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)
Khi đó pt tương đương:
\(\Leftrightarrow\frac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{6}=-x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\left(loại\right)\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
15 tháng 8 2020
3.
\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)
\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
18 tháng 7
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: \(cos7x-\sin5x=\sqrt3\left(cos5x-\sin7x\right)\)
=>\(cos7x+\sqrt3\cdot\sin7x=\sqrt3\cdot cos5x+\sin5x\)
=>\(\frac{\sqrt3}{2}\cdot\sin7x+\frac12\cdot cos7x=\frac{\sqrt3}{2}\cdot cos5x+\frac12\cdot\sin5x\)
=>\(\sin\left(7x+\frac{\pi}{6}\right)=\sin\left(5x+\frac{\pi}{3}\right)\)
=>\(\left[\begin{array}{l}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\ 7x+\frac{\pi}{6}=\pi-5x-\frac{\pi}{3}+k2\pi=-5x+\frac23\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ 12x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{12}+k\pi\\ x=\frac{1}{24}\pi+\frac{k\pi}{6}\end{array}\right.\)
