Bài 2: Phương trình lượng giác cơ bản

4C (đây là lý thuyết cơ bản trong SGK, câu này sai do dấu tương đương, nếu là đấu suy ra thì đúng)

5C (cũng là nghiệm cơ bản trong SGK)

6D (cũng SGK)

8.

\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin\left(\dfrac{\pi}{3}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k2\pi\\x=\dfrac{5\pi}{12}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

10.

\(sin\left(\dfrac{2x}{3}+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow\dfrac{2x}{3}+\dfrac{\pi}{3}=k\pi\)

\(\Leftrightarrow\dfrac{2x}{3}=-\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{2}+\dfrac{k3\pi}{2}\) (\(k\in Z\))

(Lưu ý rằng \(-k\) và \(k\) là như nhau do k là số nguyên bất kì nên bất kể trước \(...k\pi\) là đấu gì thì người ta thường chuyển hết về dấu dương)

11.

\(cosx=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow cosx=cos\left(\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{4}+k2\pi\) (\(k\in Z\))

12.

\(cosx=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) (\(k\in Z\))

13.

\(sin\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{3}+k\pi\) (\(k\in Z\))

18.

\(1-sin^22x+2\left(sinx+cosx\right)^3-3sin2x-3=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^3-\left(sin^22x+3sin2x+2\right)=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^3-\left(sin2x+1\right)\left(sin2x+2\right)=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^3+2\left(sinx+cosx\right)^2\left(sinx.cosx+1\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)^2\left(sinx+cosx+sinx.cosx+1\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)^2\left(sinx+1\right)\left(cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-cosx\\sinx=-1\\cosx=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\sinx=-1\\cosx=-1\end{matrix}\right.\)

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

19.

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

\(\Leftrightarrow\left(cosx-sinx\right)\left(1+sinx.cosx\right)=1\)

Đặt \(cosx-sinx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

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

Pt trở thành:

\(t\left(1+\dfrac{1-t^2}{2}\right)=1\)

\(\Leftrightarrow t^3-3t+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow cosx-sinx=1\)

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

\(\Leftrightarrow...\)

20.

\(sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x+1=2cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\)

\(\Leftrightarrow sin\dfrac{x}{2}.sinx-cos\dfrac{x}{2}.sin^2x-\left[2cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)-1\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}.sinx-cos\dfrac{x}{2}.sin^2x-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x-sinx=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sin\dfrac{x}{2}-cos\dfrac{x}{2}sinx-1=0\left(1\right)\end{matrix}\right.\)

Xét (1) 

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

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

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\)

\(\Leftrightarrow...\)

ĐKXĐ: \(sinx\ne0\)

Chia 2 vế cho \(sin^2x\) ta được:

\(\dfrac{3}{sin^2x}-4=4cotx.\left(\dfrac{cosx}{sinx}\right)^2-\dfrac{3cotx}{sin^2x}\)

\(\Leftrightarrow3\left(1+cot^2x\right)-4=4cotx.cot^2x-3cotx\left(1+cot^2x\right)\)

\(\Leftrightarrow3cot^2x-1=4cot^3x-3cotx-3cot^3x\)

\(\Leftrightarrow cot^3x-3cot^2x-3cotx+1=0\)

\(\Leftrightarrow\left(cotx+1\right)\left(cot^2x-4cotx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2-\sqrt{3}\\cotx=2+\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)

a.

\(sin\left(2x-\dfrac{\pi}{5}\right)=1\)

\(\Leftrightarrow2x-\dfrac{\pi}{5}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{7\pi}{20}+k\pi\)

b.

\(sin2x=\dfrac{5}{4}>1\)

\(\Rightarrow\)Phương trình đã cho vô nghiệm

c.

\(sin\left(5x-\dfrac{\pi}{3}\right)+sin2x=0\)

\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{3}\right)=-sin2x\)

\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{3}\right)=sin\left(-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{21}+\dfrac{k2\pi}{7}\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

d.

\(sin\left(7x-\dfrac{5\pi}{6}\right)=-cos\left(3x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(7x-\dfrac{5\pi}{6}\right)=sin\left(3x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-\dfrac{5\pi}{6}=3x-\dfrac{\pi}{6}+k2\pi\\7x-\dfrac{5\pi}{6}=\dfrac{7\pi}{6}-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{5}+\dfrac{k\pi}{5}\end{matrix}\right.\)

Tất cả k dưới đây là \(k\in Z\)

a.

\(2sinx=-\sqrt{2}\)

\(\Leftrightarrow sinx=-\dfrac{\sqrt{2}}{2}\)

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

b.

\(sin\left(x-2\right)=\dfrac{2}{3}\)

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

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

c.

\(sin\left(2x-\dfrac{\pi}{5}\right)=sin\left(\dfrac{\pi}{5}+x\right)\)

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

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

Lời giải:

$y=2\sin ^2x+\cos x-3$

$=2(1-\cos ^2x)+\cos x-3$

$=-2\cos ^2x+\cos x-1$

Đặt $\cos x=a$ với $a\in [-1;1]$ thì cần tìm min, max của:

$y=-2a^2+a-1$

Ta thấy: 
$y=\frac{-7}{8}-2(a-\frac{1}{4})^2\leq \frac{-7}{8}$ với mọi $a\in [-1;1]$ nên $y_{\max}=\frac{-7}{8}$

$y=(3-2a)(a+1)-4\geq -4$ với mọi $-1\leq a\leq 1$

Do đó $y_{\min}=-4$

\(y=\sqrt{5}\left(\dfrac{1}{\sqrt{5}}cosx-\dfrac{2}{\sqrt{5}}sinx\right)+3\)

Đặt \(\dfrac{1}{\sqrt{5}}=cos\alpha\Rightarrow\dfrac{2}{\sqrt{5}}=sin\alpha\)

\(\Rightarrow y=\sqrt{5}\left(cosx.cos\alpha-sinx.sin\alpha\right)+3=\sqrt{5}cos\left(x+\alpha\right)+3\)

Do \(-1\le cos\left(x+\alpha\right)\le1\)

\(\Rightarrow3-\sqrt{5}\le y\le3+\sqrt{5}\)

=>\(\left\{{}\begin{matrix}tan\left(2x-\dfrac{pi}{3}\right)=\sqrt{2}\\tan\left(2x-\dfrac{pi}{3}\right)=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{pi}{3}=arctan\left(\sqrt{2}\right)+kpi\\2x-\dfrac{pi}{3}=arctan\left(-\sqrt{2}\right)+kpi\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(\dfrac{pi}{3}+arctan\left(\sqrt{2}\right)+kpi\right)\\x=\dfrac{1}{2}\left(\dfrac{pi}{3}+arctan\left(-\sqrt{2}\right)+kpi\right)\end{matrix}\right.\)

`2cos x+1=0`

`<=>cos x=-1/2`

`<=>x=[+-2\pi]/3+k2\pi`   `(k in ZZ)`

\(2\cos x+1=0\)

\(\Leftrightarrow2\cos x=-1\)

\(\Leftrightarrow\cos x=-\dfrac{1}{2}\)

\(\Leftrightarrow\cos x=\cos\left(\dfrac{2\pi}{3}\right)\)

\(\Leftrightarrow x=\pm\dfrac{2\pi}{3}+k2\pi\) \((k\in \mathbb Z)\)

`a)sin(1-x)=\sqrt{3}/2`

`<=>[(1-x=\pi/3+k2\pi),(1-x=[2\pi]/3+k2\pi):}`

`<=>[(x=1-\pi/3-k2\pi),(x=1-[2\pi]/3-k2\pi):}`      `(k in ZZ)`

__________________________

`b->` Bạn vt lại đề nhé