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Câu 1:
\(\Leftrightarrow sinx.cos\frac{\pi}{3}-cosx.sin\frac{\pi}{3}+2\left(cosx.cos\frac{\pi}{6}+sinx.sin\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow sinx+\frac{1}{\sqrt{3}}cosx=0\)
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cosx\)
\(tanx+\frac{1}{\sqrt{3}}=0\Rightarrow tanx=-\frac{1}{\sqrt{3}}\Rightarrow x=\frac{\pi}{6}+k\pi\)
Câu 2:
\(\Leftrightarrow1-cos6x=1+cos2x\)
\(\Leftrightarrow-cos6x=cos2x\)
\(\Leftrightarrow cos\left(\pi-6x\right)=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi-6x+k2\pi\\2x=6x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
Câu 3:
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}-4\pi\right)+cos2x=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)+cos2x=1\)
\(\Leftrightarrow cos2x+cos2x=1\)
\(\Leftrightarrow cos2x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Câu 4:
\(\sqrt{2}\left(cosx.cos\frac{3\pi}{4}+sinx.sin\frac{3\pi}{4}\right)=1+sinx\)
\(\Leftrightarrow-cosx+sinx=1+sinx\)
\(\Leftrightarrow cosx=-1\Rightarrow x=\pi+k\pi2\)
Câu 5:
Giống câu 3, chắc bạn ghi nhầm đề
a.
\(\Leftrightarrow2sin\frac{17\pi}{30}cos\left(3x-\frac{7\pi}{30}\right)=\sqrt{3}\)
\(\Leftrightarrow cos\left(3x-\frac{7\pi}{30}\right)=\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}\)
Đặt \(\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow cos\left(3x-\frac{7\pi}{30}\right)=cosa\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{7\pi}{30}=a+k2\pi\\3x-\frac{7\pi}{30}=-a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{90}+\frac{a}{3}+\frac{k2\pi}{3}\\x=\frac{7\pi}{30}-\frac{a}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
Chắc bạn ghi sai đề, con số \(\frac{4\pi}{3}\) sẽ hợp lý hơn con số \(\frac{4\pi}{5}\) rất nhiều
Với \(sinx=0\) không phải nghiệm (vế trái bằng 9, vế phải hiển nhiên nhỏ hơn 9)
Với \(sinx\ne0\):
\(\Rightarrow\left(3sinx-4sin^3x\right)\left(3-4sin^23x\right)=sinx-2sinx.cos10x\)
\(\Leftrightarrow sin3x\left(3-4sin^23x\right)=sinx-2sinx.cos10x\)
\(\Leftrightarrow3sin3x-4sin^33x=sinx-sin11x+sin9x\)
\(\Leftrightarrow sin9x=sinx-sin11x+sin9x\)
\(\Leftrightarrow sin11x=sinx\)
\(\Leftrightarrow...\)
ĐKXĐ: ...
\(\Leftrightarrow tan^2x+cot^2x=2\left(cos^4x+sin^4x+2sin^2x.cos^2x\right)\)
\(\Leftrightarrow tan^2x+cot^2x=2\left(sin^2x+cos^2x\right)^2\)
\(\Leftrightarrow tan^2x+cot^2x=2\)
\(\Leftrightarrow\left(tanx-cotx\right)^2=0\)
\(\Leftrightarrow tanx=cotx=tan\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow x=\frac{\pi}{2}-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
a)\(pt\Leftrightarrow\frac{1-cos8x}{2}+\frac{1-cos6x}{2}=\frac{1-cos4x}{2}+\frac{1-cos2x}{2}\)
\(\Leftrightarrow cos2x+cos4x=cos6x+cos8x\)
\(\Leftrightarrow2cos3x\cdot cosx=2cos7x\cdot cosx\)
\(\Leftrightarrow2cos\left(cos3x-cos7x\right)=0\)
\(\Leftrightarrow2cosx\cdot\left(-2\right)\cdot sin5x\cdot sin\left(-2x\right)=0\)
\(\Leftrightarrow cosx\cdot sin2x\cdot sin5x=0\)
\(\Leftrightarrow sin2x\cdot sin5x=0\)(do sin2x=0 <=>2sinx*cosx=0 gồm th cosx=0 r`)
\(\Leftrightarrow\left[\begin{array}{nghiempt}sin2x=0\\sin5x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{k\pi}{2}\\x=\frac{k\pi}{5}\end{array}\right.\)\(\left(k\in Z\right)\)
b)\(pt\Leftrightarrow1-cos2x+1-cos4x=1+cos6x+1+cos8x\)
\(\Leftrightarrow cos2x+cos8x+cos4x+cos6x=0\)
\(\Leftrightarrow cos10x\cdot cos6x+cos10x\cdot cos2x=0\)
\(\Leftrightarrow cos10x\left(cos6x+cos2x\right)=0\)
\(\Leftrightarrow cos10x\cdot cos8x\cdot cos4x=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}cos10x=0\\cos8x=0\\cos4x=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{16}+\frac{k\pi}{8}\\x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)