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)

KL:....

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

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

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

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

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

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

Đáp án đúng : A

a: =>2sin(x+pi/3)=-1

=>sin(x+pi/3)=-1/2

=>x+pi/3=-pi/6+k2pi hoặc x+pi/3=7/6pi+k2pi

=>x=-1/2pi+k2pi hoặc x=2/3pi+k2pi

b: =>2sin(x-30 độ)=-1

=>sin(x-30 độ)=-1/2

=>x-30 độ=-30 độ+k*360 độ hoặc x-30 độ=180 độ+30 độ+k*360 độ

=>x=k*360 độ hoặc x=240 độ+k*360 độ

c: =>2sin(x-pi/6)=-căn 3

=>sin(x-pi/6)=-căn 3/2

=>x-pi/6=-pi/3+k2pi hoặc x-pi/6=4/3pi+k2pi

=>x=-1/6pi+k2pi hoặc x=3/2pi+k2pi

d: =>2sin(x+10 độ)=-căn 3

=>sin(x+10 độ)=-căn 3/2

=>x+10 độ=-60 độ+k*360 độ hoặc x+10 độ=240 độ+k*360 độ

=>x=-70 độ+k*360 độ hoặc x=230 độ+k*360 độ

e: \(\Leftrightarrow2\cdot sin\left(x-15^0\right)=-\sqrt{2}\)

=>\(sin\left(x-15^0\right)=-\dfrac{\sqrt{2}}{2}\)

=>x-15 độ=-45 độ+k*360 độ hoặc x-15 độ=225 độ+k*360 độ

=>x=-30 độ+k*360 độ hoặc x=240 độ+k*360 độ

f: \(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=-\dfrac{1}{\sqrt{2}}\)

=>x-pi/3=-pi/4+k2pi hoặc x-pi/3=5/4pi+k2pi

=>x=pi/12+k2pi hoặc x=19/12pi+k2pi

g) \(3+\sqrt[]{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=-\dfrac{3}{\sqrt[]{5}}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=sin\left[arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)\right]\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\\x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

h) \(1+sin\left(x-30^o\right)=0\)

\(\Leftrightarrow sin\left(x-30^o\right)=-1\)

\(\Leftrightarrow sin\left(x-30^o\right)=sin\left(-90^o\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^o=-90^0+k360^o\\x-30^o=180^o+90^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-60^0+k360^o\\x=300^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow x=-60^0+k360^o\)

\(A=\frac{3sina-2cosa}{12sin^3a+4cos^3a}=\frac{\frac{3sina}{sin^3a}-\frac{2cosa}{sin^3a}}{12+\frac{4cos^3a}{sin^3a}}=\frac{3.\frac{1}{sin^2a}-2cota.\frac{1}{sin^2a}}{12+4cot^3a}\)

\(=\frac{3\left(1+cot^2a\right)-2cota\left(1+cot^2a\right)}{12+4cot^3a}=\frac{3\left(1+3^2\right)-2.3.\left(1+3^2\right)}{12+4.3^3}=...\)

Câu a)

Đặt \(2x=a\). PT trở thành:

\(2\sin ^2a+\sin 3a-1=\sin a\)

\(\Leftrightarrow 2\sin ^2a+\sin (a+2a)-1-\sin a=0\)

\(\Leftrightarrow 2\sin ^2a+\sin a\cos 2a+\cos a\sin 2a-1-\sin a=0\)

\(\Leftrightarrow 2\sin ^2a+\sin a\cos 2a+2\cos ^2a\sin a-1-\sin a=0\)

\(\Leftrightarrow (2\sin ^2a-1)+\sin a\cos 2a+\sin a(2\cos ^2a-1)=0\)

\(\Leftrightarrow -\cos 2a+\sin a\cos 2a+\sin a\cos 2a=0\)

\(\Leftrightarrow \cos 2a(-1+2\sin a)=0\)

\(\Rightarrow \left[\begin{matrix} \cos 2a=0(1)\\ \sin a=\frac{1}{2}(2)\end{matrix}\right.\)

Từ (1) \(\Rightarrow 2a=\frac{\pi}{2}+k\pi (k\in\mathbb{Z})\)\(\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)

Từ (2) \(\Rightarrow \left[\begin{matrix} a=\frac{\pi}{6}+2k\pi \rightarrow x=\frac{\pi}{12}+k\pi \\ a=\frac{5}{6}\pi+2k\pi \rightarrow x=\frac{5\pi}{12}+k\pi \end{matrix}\right.\)

Bài 2:

\(\sin 2x+\sin 6x+2\sin ^2x-1=0\)

\(\Leftrightarrow \sin 2x+\sin 6x-\cos 2x=0\)

\(\Leftrightarrow \sin 2x+\sin 4x\cos 2x+\cos 4x\sin 2x-\cos 2x=0\)

\(\Leftrightarrow \sin a+\sin 2a\cos a+\cos 2a\sin a-\cos a=0\)

\(\Leftrightarrow \sin a(1+\cos 2a)+\sin 2a\cos a-\cos a=0\)

\(\Leftrightarrow \sin a.2\cos ^2a+\sin 2a\cos a-\cos a=0\)

\(\Leftrightarrow \cos a(2\sin 2a-1)=0\)

\(\Rightarrow \left[\begin{matrix} \cos a=0(1)\\ \sin 2a=\frac{1}{2}(2)\end{matrix}\right.\)

Từ (1)\(\Rightarrow a=\frac{\pi}{2}+k\pi \Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

Từ (2) \(\Rightarrow \left[\begin{matrix} 2a=\frac{\pi}{6}+2k\pi \rightarrow x=\frac{\pi}{24}+\frac{k\pi}{2}\\ 2a=\frac{5\pi}{6}+2k\pi \rightarrow x=\frac{5\pi}{24}+\frac{k\pi}{2}\end{matrix}\right.\)

lm trên symbolab.com

\(\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\cos^2x\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\left(2-\sin^2x\right)\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=4\sin^2x-1\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=\left(2\sin x-1\right)\left(2\sin x+1\right)\)

\(\Leftrightarrow2\sin2x+1=2\sin x+1\)

\(\Leftrightarrow\sin2x=\sin x\)

\(\Leftrightarrow\sin2x-\sin x=0\)

\(\Leftrightarrow2\cos\frac{3}{2}-\cos\frac{x}{2}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\cos\frac{3}{2}=0\\\cos\frac{x}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{cases}\left(k\inℤ\right)}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{3}+\frac{2\pi}{3}k\\x=\pi+4k\pi\end{cases}\left(k\inℤ\right)}\)

Đề thiếu. Bạn xem lại đề.