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\(\Leftrightarrow2x+\frac{\pi}{3}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{2}\)
Do \(x\in\left[0;2\pi\right]\Rightarrow0\le\frac{\pi}{12}+\frac{k\pi}{2}\le2\pi\)
\(\Rightarrow-\frac{1}{6}\le k\le\frac{23}{6}\Rightarrow k=\left\{0;1;2;3\right\}\)
\(\Rightarrow x=\left\{\frac{\pi}{12};\frac{7\pi}{12};\frac{13\pi}{12};\frac{19\pi}{12}\right\}\)
\(sin\left(x+\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=-\frac{\pi}{3}+k2\pi\\x+\frac{\pi}{6}=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
Do \(x\in\left[0;2\pi\right]\Rightarrow\left[{}\begin{matrix}0\le-\frac{\pi}{2}+k2\pi\le2\pi\\0\le\frac{7\pi}{6}+k2\pi\le2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}k=1\\k=0\end{matrix}\right.\) \(\Rightarrow x=\left\{\frac{3\pi}{2};\frac{7\pi}{6}\right\}\)
\(cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=-\pi+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\pi\right\}\)
\(cos2x=-1\Leftrightarrow2x=\pi+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{2}+k\pi\)
\(x\in\left[0;2\pi\right]\Rightarrow0\le\frac{\pi}{2}+k\pi\le2\pi\)
\(\Rightarrow-\frac{1}{2}\le k\le\frac{3}{2}\Rightarrow k=\left\{0;1\right\}\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\frac{3\pi}{2}\right\}\)
\(\Leftrightarrow3\sin x-4\sin^3x+4\cos^3x-3\cos x-2\cos x+2\sin x+1=0\)\(\Leftrightarrow4\left[\left(\cos x-\sin x\right)^3+3\cos x.\sin x\left(\cos x-\sin x\right)\right]-5\left(\cos x-\sin x\right)+1=0\)\(\Leftrightarrow4\left[\left(\cos x-\sin x\right)^3+3\dfrac{\left(\cos x-\sin x\right)^2-1}{2}\left(\cos x-\sin x\right)\right]-5\left(\cos x-\sin x\right)+1=0\)Đặt cosx-sinx=a. Thay vào giải pt ta tìm được: a=1
<=> cosx-sinx=1
\(\Leftrightarrow\cos x.\sin\dfrac{\pi}{4}-\sin x.\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\sin\left(\dfrac{\pi}{4}-x\right)=\sin\dfrac{\pi}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{4}-x=\dfrac{\pi}{4}-2k\pi\Rightarrow x=2k\pi\\\dfrac{\pi}{4}-x=\pi-\dfrac{\pi}{4}-2k\pi\Rightarrow x=-\dfrac{\pi}{2}+2k\pi\end{matrix}\right.\)
a) Pt\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x\left(sin^2x+cos^2x\right)+3sinx.cosx-\dfrac{m}{4}+2=0\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x-\dfrac{3}{2}sin2x-\dfrac{m}{4}+2=0\)
\(\Leftrightarrow-3sin^22x-6sin2x-m+12=0\)
Đặt \(t=sin2x;t\in\left[-1;1\right]\)
Pttt: \(-3t^2-6t-m+12=0\)
\(\Leftrightarrow-3t^2-6t+12=m\) (1)
Đặt \(f\left(t\right)=-3t^2-6t+12;t\in\left[-1;1\right]\)
Vẽ BBT sẽ tìm được \(f\left(t\right)_{min}=3;f\left(t\right)_{max}=15\)\(\Leftrightarrow3\le f\left(t\right)\le15\)\(\Rightarrow m\in\left[3;15\right]\) thì pt (1) sẽ có nghiệm
mà \(m\in Z\) nên tổng m nguyên để pt có nghiệm là 13 m
Vậy có tổng 13 m nguyên
b) Pt\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(1\right)\\2cos^2x-\left(2m+1\right)cosx+m=0\left(2\right)\end{matrix}\right.\)
Từ (1)\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)
\(x\in\left[0;2\pi\right]\Rightarrow0\le\dfrac{\pi}{2}+k2\pi\le2\pi\)\(\Leftrightarrow-\dfrac{1}{4}\le k\le\dfrac{3}{4}\)\(\Rightarrow k=0\)
Tại k=0\(\Rightarrow x=\dfrac{\pi}{2}\)
Để pt ban đầu có 4 nghiệm pb \(\in\left[0;2\pi\right]\)
\(\Leftrightarrow\) Pt (2) có 3 nghiệm pb khác \(\dfrac{\pi}{2}\)
Xét pt (2) có: \(2cos^2x-\left(2m+1\right)cosx+m=0\)
Vì là phương trình bậc hai ẩn \(cosx\) nên pt (2) chỉ có nhiều nhất ba nghiệm \(\Leftrightarrow\) Pt (2) có một nghiệm cosx=0
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) mà \(x\ne\dfrac{\pi}{2}\)
\(\Rightarrow\) Pt (2) chỉ có nhiều nhất hai nghiệm
\(\Rightarrow\) Pt ban đầu không thể có 4 nghiệm phân biệt
Vậy \(m\in\varnothing\)
\(cosx-\left(3sinx-4sin^3x\right)=\sqrt{2}\left(cosx-sinx\right)sin4x\)
\(\Leftrightarrow cosx-sinx+2sinx\left(2sin^2x-1\right)=\sqrt{2}\left(cosx-sinx\right)sin4x\)
\(\Leftrightarrow cosx-sinx-2sinx\left(cosx-sinx\right)\left(cosx+sinx\right)=\sqrt{2}\left(cosx-sinx\right)sin4x\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(1-2sinx\left(sinx+cosx\right)-\sqrt{2}sin4x\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(1-2sin^2x-2sinx.cosx-\sqrt{2}sin4x\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cos2x-sin2x-\sqrt{2}sin4x=0\right)\)
\(\Leftrightarrow\left(cosx-sinx\right)\left[sin\left(\dfrac{\pi}{4}-2x\right)-sin4x\right]=0\)
\(\Leftrightarrow...\)
Đặt \(x+\dfrac{\pi}{6}=t\Rightarrow x=t-\dfrac{\pi}{6}\Rightarrow3x=3t-\dfrac{\pi}{2}\)
\(2cost=sin\left(3t-\dfrac{\pi}{2}\right)-cos\left(3t-\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow2cost=-cos3t-sin3t\)
\(\Leftrightarrow2cost=3cost-4cos^3t+4sin^3t-3sint\)
\(\Leftrightarrow4sin^3t-3sint+cost-4cos^3t=0\)
\(cost=0\) không phải nghiệm
\(\Rightarrow4tan^3t-3tant\left(1+tan^2t\right)+1+tan^2t-4=0\)
\(\Leftrightarrow tan^3t+tan^2t-3tant-3=0\)
\(\Leftrightarrow\left(tant+1\right)\left(tan^2t-3\right)=0\)
\(\Leftrightarrow...\)
\(\Leftrightarrow sin2x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\frac{\pi}{3}+k2\pi\\2x=\frac{2\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
\(0< x< 2\pi\Rightarrow\left[{}\begin{matrix}0< \frac{\pi}{3}+k2\pi< 2\pi\\0< \frac{2\pi}{9}+\frac{k2\pi}{3}< 2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\frac{1}{6}< k< \frac{5}{6}\\-\frac{1}{3}< k< \frac{8}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}k=0\\k=\left\{0;1;2\right\}\end{matrix}\right.\) \(\Rightarrow x=\left\{\frac{\pi}{3};\frac{2\pi}{9};\frac{8\pi}{9};\frac{14\pi}{9}\right\}\)
\(sin3x=0\Leftrightarrow3x=k\pi\)
\(\Leftrightarrow x=\frac{k\pi}{3}\)
Do \(x\in\left[0;2\pi\right]\Rightarrow0\le\frac{k\pi}{3}\le2\pi\Rightarrow0\le k\le6\)
\(\Rightarrow x=\left\{0;\frac{\pi}{3};\frac{2\pi}{3};\pi;\frac{4\pi}{3};\frac{5\pi}{3};2\pi\right\}\)