Question

[Giang]

Giải chi tiết giúp mình với 

Answer 1

[Nguyễn Việt Lâm]

a.

Công thức góc cơ bản: \(cos\left(a+k\pi\right)=\pm cosa\) ; \(sin\left(a+k2\pi\right)=sina\) ; \(cos\left(a+\dfrac{\pi}{2}\right)=-sina\)

Do đó pt tương đương:

\(2cosx+\dfrac{1}{3}cos^2x=\dfrac{8}{3}+sin2x-3sinx+\dfrac{1}{3}sin^2x\)

\(\Leftrightarrow6cosx+1-sin^2x=8+3sin2x-9sinx+sin^2x\)

\(\Leftrightarrow2sin^2x-9sinx+7+6sinx.cosx-6cosx=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(2sinx-7\right)+6cosx\left(sinx-1\right)=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(2sinx+6cosx-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\\2sinx+6cosx=7\left(1\right)\end{matrix}\right.\)

Xét (1), ta có \(2^2+6^2=40< 7^2\) nên (1) vô nghiệm

Vậy họ nghiệm của pt là \(x=\dfrac{\pi}{2}+k2\pi\)

Answer 2

[Nguyễn Việt Lâm]

b.

Ta có:

\(tan^2x\left(1-sin^3x\right)+cos^3x-1=0\)

\(\Leftrightarrow\dfrac{\left(1-cos^2x\right)\left(1-sin^3x\right)}{\left(1-sin^2x\right)}-\left(1-cos^3x\right)=0\)

\(\Leftrightarrow\dfrac{\left(1-cosx\right)\left(1+cosx\right)\left(1+sinx+sin^2x\right)}{1+sinx}-\left(1-cosx\right)\left(1+cosx+cos^2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Rightarrow x=k2\pi\\\dfrac{\left(1+cosx\right)\left(1+sinx+sin^2x\right)}{1+sinx}=1+cosx+cos^2x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left(1+cosx\right)\left(1+sinx+sin^2x\right)=\left(1+sinx\right)\left(1+cosx+cos^2x\right)\)

\(\Leftrightarrow sin^2x+sin^2x.cosx=cos^2x+cos^2x.sinx\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+cosx\right)+sinx.cosx\left(sinx-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\Rightarrow x=\dfrac{\pi}{4}+k\pi\\sinx+cosx+sinx.cosx=0\left(2\right)\end{matrix}\right.\)

Xét (2), đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)

\(sinx.cosx=\dfrac{t^2-1}{2}\)

Pt (2) trở thành:

\(t+\dfrac{t^2-1}{2}=0\Leftrightarrow t^2+2t-1=0\Rightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(loại\right)\end{matrix}\right.\)

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

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}-1}{\sqrt{2}}\)

\(\Rightarrow...\)