2cox2x-1+12(1-cox2x/2)-1=0 <=> 2cox²2x-1+6-6cox2x-1=0 <=> 2cox²2x-6cox2x+4=0   <=> cos2x=1 (nhận) hoặc cox2x=2 (loại)

<=> 2x=k2π

<=>×=kπ (k£Z)

Chọn đáp án C

a, Ta có:  sin 4 x + cos 4 x = sin 2 x + cos 2 x 2 - 2 sin 2 x . cos 2 x = 1 - 2 sin 2 x . cos 2 x

b, Ta có:  sin 6 x + cos 6 x = sin 2 x + cos 2 x 3 - 3 sin 2 x cos 2 x sin 2 x + cos 2 x =  1 - 3 sin 2 x cos 2 x

a: cos5x=-5

mà -1<=cos5x<=1

nên \(x\in\varnothing\)

b: 2*cosx-1=0

=>2*cosx=1

=>cosx=1/2

=>x=pi/3+k2pi hoặc x=-pi/3+k2pi

c: -5*cos(x+pi/3)=0

=>cos(x+pi/3)=0

=>x+pi/3=pi/2+kpi

=>x=pi/6+kpi

d: cos4x=cos(5/12pi)

=>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi

=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2

e: cos^2x=1

=>sin^2x=0

=>sin x=0

=>x=kpi

\(cos4x\cdot\sqrt{\dfrac{\pi^2}{9}-x^2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\\sqrt{\dfrac{\pi^2}{9}-x^2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\\\dfrac{\pi^2}{9}-x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\left(k\in Z\right)\\x=\pm\dfrac{\pi}{3}\end{matrix}\right.\)

\(tanx=\dfrac{1}{cotx}=\dfrac{1}{\sqrt[]{2}}=\dfrac{\sqrt[]{2}}{2}\left(tanx.cotx=1\right)\)

\(1+tan^2x=\dfrac{1}{cos^2x}\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}=\dfrac{1}{1+\dfrac{1}{2}}\)

\(\Rightarrow cos^2x=\dfrac{2}{3}\Rightarrow cosx=\sqrt[]{\dfrac{2}{3}}\)

\(tanx=\dfrac{sinx}{cosx}\Rightarrow sinx=tanx.cosx=\dfrac{1}{\sqrt[]{2}}.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{3}\)

\(P=\dfrac{3sinx-2cosx}{12sin^3x+4cos^3x}=\dfrac{3.\dfrac{\sqrt[]{3}}{3}-2.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)

\(=\dfrac{\sqrt[]{3}-\dfrac{2\sqrt[]{6}}{3}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)

Đáp án đúng : A

Đề là thế này: \(1+cos4x-2sin^2x=0\)

Hay thế này: \(1+cos4x-2sin2x=0\)

\(1+cos4x-2sin2x=0\)

\(\Leftrightarrow1+1-2sin^22x-2sin2x=0\)

\(\Leftrightarrow sin^22x+sin2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{5}-1}{2}=sin\alpha\\sin2x=\frac{-\sqrt{5}-1}{2}< -1\left(l\right)\end{matrix}\right.\)

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

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