1: \(3\cdot\sin^23x+4\cdot\sin6x+\left(8\sqrt3-9\right)\cdot cos^23x=0\) (1)

TH1: cos3x=0

=>\(sin^23x=1\) và sin 6x=0

(1) sẽ trở thành: 3*1+4*0+(8\(\sqrt3-9\) )*0=0

=>3=0(vô lý)

=>Loại

Th2: cos3x<>0

Từ (1) suy ra \(3\cdot\frac{\sin^23x}{cos^23x}+4\cdot\frac{\sin6x}{cos^23x}+\left(8\sqrt3-9\right)\cdot cos^23x:cos^23x=0\)

=>\(3\cdot\tan^23x+8\cdot\tan3x+8\sqrt3-9=0\) (2)

\(\Delta=8^2-4\cdot3\cdot\left(8\sqrt3-9\right)=64-96\sqrt3+108=172-96\sqrt3\)

\(=4\left(43-24\sqrt3\right)=4\left(3\sqrt3-4\right)^2\)

(2) sẽ có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}\tan3x=\frac{-8-2\cdot\left(3\sqrt3-4\right)}{2\cdot3}=\frac{-8-6\sqrt3+8}{6}=-\sqrt3\\ \tan3x=\frac{-8+2\left(3\sqrt3-4\right)}{2\cdot3}=\frac{-8+6\sqrt3-8}{6}=\frac{6\sqrt3-16}{6}=\frac{3\sqrt3-8}{3}\end{array}\right.\)

\(\tan3x=-\sqrt3\)

=>\(3x=-\frac{\pi}{3}+k\pi\)

=>\(x=-\frac{\pi}{9}+\frac{k\pi}{3}\)

\(\tan3x=\frac{3\sqrt3-8}{3}\)

=>\(3x=\arctan\left(\frac{3\sqrt3-8}{3}\right)+k\pi\)

=>\(x=\frac13\cdot\arctan\left(\frac{3\sqrt3-8}{3}\right)+\frac{k\pi}{3}\)

\(pt\Leftrightarrow cos6x+3cos2x-4\left(2cos^2x-1\right)=0\)

\(\Leftrightarrow cos6x+3cos2x-4cos2x=0\)

\(\Leftrightarrow cos6x-cos2x=0\)

\(\Leftrightarrow-2sin4x.sin2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin2x=0\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{k\pi}{4}\)

Chứng minh các biểu thức đã cho không phụ thuộc vào x.

f(x) = 1 ⇒ f′(x) = 0

2sin^2(2x+pi/3)-6sin(x+pi/6)+cos(x+pi/6)+2=0

e/

ĐKXĐ: ...

\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)

\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)

\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)

d/

Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)

\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)