Bài 1: Hàm số lượng giác

2.1

a, \(sin3x=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin3x=sin\left(-\dfrac{\pi}{3}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

b, \(sin\left(2x-15^o\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-15^o\right)=sin45^o\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-15^o=45^o+k360^o\\2x-15^o=135^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30^o+k180^o\\x=75^o+k180^o\end{matrix}\right.\)

2.1

c, \(sin\left(\dfrac{x}{2}+10^o\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(\dfrac{x}{2}+10^o\right)=sin\left(-30^o\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}+10^o=-30^o+k360^o\\\dfrac{x}{2}+10^o=210^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-40^o+k720^o\\x=400^o+k720^o\end{matrix}\right.\)

d, \(sin4x=\dfrac{2}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=arcsin\dfrac{2}{3}+k2\pi\\4x=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{arcsin\dfrac{2}{3}}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi-arcsin\dfrac{2}{3}}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

2.2

a, \(cos\left(x+3\right)=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=arccos\dfrac{1}{3}+k2\pi\\x+3=\pi-arccos\dfrac{1}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arccos\dfrac{1}{3}-3+k2\pi\\x=\pi-arccos\dfrac{1}{3}-3+k2\pi\end{matrix}\right.\)

b, \(cos\left(3x-45^o\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow cos\left(3x-45^o\right)=cos30^o\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-45^o=30^o+k360^o\\3x-45^o=150^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=25^o+k120^o\\x=65^o+k120^o\end{matrix}\right.\)

Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\)

Khi đó \(y=f\left(t\right)=t^2-4t+5\)

\(M=maxf\left(t\right)=max\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(-1\right)=10\)

\(m=min\left(t\right)=min\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(1\right)=2\)

\(\Rightarrow P=M-2m^2=10-2.4=2\)

a) Hàm số xđ <=> \(1+cos2x>0\) \(\Leftrightarrow cos2x\ne-1\) \(\Leftrightarrow\)\(2cos^2x-1\ne-1\)

\(\Leftrightarrow cosx\ne0\) \(\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

b)Hàm số xđ <=> \(1-sinx>0\) \(\Leftrightarrow sinx\ne1\) \(\Leftrightarrow x\ne\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)

c) Hàm số xđ <=> \(sinx+cos5x\ne0\)

\(\Leftrightarrow sinx\ne-cos5x\)

\(\Leftrightarrow cos\left(\dfrac{\pi}{2}-x\right)\ne cos\left(\pi-5x\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\pi}{2}-x\ne\pi-5x+k2\pi\\\dfrac{\pi}{2}-x\ne-\pi+5x+k2\pi\end{matrix}\right.\) (\(k\in Z\))

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{4}-\dfrac{k\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

d) Hàm số xđ <=> \(sinx-\sqrt{3}cosx\ne0\)

\(\Leftrightarrow2.sin\left(x-\dfrac{\pi}{3}\right)\ne0\) \(\Leftrightarrow x\ne\dfrac{\pi}{3}+k\pi\left(k\in Z\right)\)

e) Hàm số xđ <=> \(\left(sinx+1\right).cosx\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne-1\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{2}+k2\pi\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) (\(k\in Z\)) \(\Rightarrow x\ne\dfrac{\pi}{2}+k\pi\) (Hai họ nghiệm trùng nhau nên e tổng hợp lại, e nghĩ thế)

f) Hàm số xđ <=> \(\left\{{}\begin{matrix}\left(1-tanx\right)\left(1-cotx\right)\ne0\\sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}tanx\ne1\\cotx\ne1\\sinx.cosx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne cosx\\\dfrac{1}{2}.sin2x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne sin\left(\dfrac{\pi}{2}-x\right)\\2x\ne k\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}-x+k2\pi\\x\ne\dfrac{\pi}{2}+x+k2\pi\\x\ne\dfrac{k\pi}{2}\end{matrix}\right.\)(\(k\in Z\))

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\0\ne\dfrac{\pi}{2}+k2\pi\\x\ne\dfrac{k\pi}{2}\end{matrix}\right.\)(\(k\in Z\)) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{k\pi}{2}\end{matrix}\right.\)(\(k\in Z\))

1) Nhận xét: \(1-cosx\ge0\forall x\) ; \(1+sinx\ge0\forall x\)

Hàm số xđ <=> \(\left\{{}\begin{matrix}\dfrac{1-cosx}{1+sinx}\ge0\left(lđ\right)\\1+sinx\ne0\end{matrix}\right.\) \(\Rightarrow sinx\ne-1\) \(\Leftrightarrow x\ne\dfrac{-\pi}{2}+k2\pi\left(k\in Z\right)\)

2) Hàm số xđ <=> \(\left\{{}\begin{matrix}tanx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\)\(\Rightarrow sinx.cosx\ne0\) 

\(\Leftrightarrow\dfrac{1}{2}.sin2x\ne0\) \(\Leftrightarrow x\ne\dfrac{k\pi}{2}\left(k\in Z\right)\)

3)Hàm số xđ <=>\(1-sin\left(x^2+2x-1\right)\ge0\)

\(\Leftrightarrow sin\left(x^2+2x-1\right)\le1\) (lđ với mọi x)

Vậy tập xđ là D=R

4)Hàm số xđ <=> \(1-cosx\ne0\) \(\Leftrightarrow cosx\ne1\) \(\Leftrightarrow x\ne k2\pi\left(k\in Z\right)\)

6) Hàm số xđ <=> \(\left\{{}\begin{matrix}sinx\ne0\\cos2x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\2cos^2x-1\ne0\end{matrix}\right.\)(\(k\in Z\)) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\cosx\ne\pm\sqrt{\dfrac{1}{2}}\end{matrix}\right.\)(\(k\in Z\))

\(\left\{{}\begin{matrix}x\ne k\pi\\x\ne\pm\dfrac{\pi}{4}+k2\pi\\x\ne\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

a. Do \(sinx\le1\) ; \(\forall x\Rightarrow3-sinx>0\) ; \(\forall x\)

\(\Rightarrow\) Hàm số xác định trên R

b. ĐKXĐ: \(sinx\ne0\Rightarrow x\ne k\pi\)

c. ĐKXĐ: \(1+cosx\ne0\Leftrightarrow cosx\ne-1\Rightarrow x\ne\pi+k2\pi\)

d. ĐKXĐ: \(cos\left(2x+\dfrac{\pi}{3}\right)\ne0\Leftrightarrow2x+\dfrac{\pi}{3}\ne\dfrac{\pi}{2}+k\pi\)

\(\Rightarrow x\ne\dfrac{\pi}{12}+\dfrac{k\pi}{2}\)

Ta có: u₈ = u₁ + 7d

⇒ 16 = -5 + 7d

⇒ d = 3