Bài 2: Phương trình lượng giác cơ bản - Hỏi đáp

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{29\pi}{126}+\frac{k2\pi}{3}\\x=\frac{41\pi}{126}+\frac{k2\pi}{3}\end{matrix}\right.\)

a/

\(sin^2x-sinx=2\left(1-sin^2x\right)\)

\(\Leftrightarrow3sin^2x-sinx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{2}{3}\end{matrix}\right.\)

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

2.

\(2sin^2x+\left(1-\sqrt{3}\right)sinx-\frac{\sqrt{3}}{2}=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

3.

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{4}=\frac{\pi}{8}+k2\pi\\3x+\frac{\pi}{4}=-\frac{\pi}{8}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+\frac{k2\pi}{3}\\x=-\frac{\pi}{8}+\frac{k2\pi}{3}\end{matrix}\right.\)

e/ Tử số đến đâu và mẫu số đến đâu bạn?

f/ Căn đến đâu bạn?

g/ Căn đến đâu bạn?

h/ \(y=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)

Do \(0\le sin^22x\le1\Rightarrow\frac{1}{2}\le y\le1\)

\(y_{max}=1\) khi \(sin^22x=0\)

\(y_{min}=\frac{1}{2}\) khi \(sin^22x=1\)

t/ \(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)

\(y=1-3sin^2x.cos^2x=1-\frac{3}{4}\left(2sinx.cosx\right)^2\)

\(y=1-\frac{3}{4}sin^22x\)

Tượng tự câu trên \(\Rightarrow\frac{1}{4}\le y\le1\)

\(y_{min}=\frac{1}{4}\) khi \(sin^22x=1\)

\(y_{max}=1\) khi \(sin^22x=0\)

Tốt nhất là bạn sử dụng công cụ gõ công thức

\(\sin^6x+\cos^6x=\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\)

\(=\left(\sin^2x+\cos^2x\right)\left(\sin^4x-\sin^2x.\cos^2x+\cos^4x\right)\)

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

\(\Rightarrow-3\sin^2x\cos^2x=\frac{15}{8}.\left(2\cos^2x-1\right)-\frac{1}{2}\)

\(\Leftrightarrow-3\sin^2x\cos^2x=\frac{15}{4}\cos^2x-\frac{15}{8}-\frac{1}{2}\)

\(\Leftrightarrow\frac{5}{4}\cos^2x+\left(1-\cos^2x\right)\cos^2x-\frac{19}{24}=0\)

\(\Leftrightarrow\cos^4x-\frac{9}{4}\cos^2x+\frac{19}{24}=0\)

Này là phương trình trùng phương, đặt \(\cos^2x=t\ge0\) rồi giải như bình thường thôi bạn

27.

\(\cos\left(\frac{\pi}{6}-2x\right)=\sin x\)

\(\Leftrightarrow\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+2x\right)=\sin x\)

\(\Leftrightarrow\sin\left(\frac{\pi}{3}+2x\right)=\sin x\)

\(\Leftrightarrow\frac{\pi}{3}+2x=\pi-x+k2\pi\Leftrightarrow x=\frac{2}{9}\pi+\frac{2}{3}k\pi\)

\(\frac{\pi}{2}< \frac{2}{9}\pi+\frac{2}{3}k\pi< \pi\Leftrightarrow\frac{5}{18}\pi< \frac{2}{3}k\pi< \frac{7}{9}\pi\)

\(\Leftrightarrow\frac{5}{12}< k< \frac{7}{6}\Rightarrow k=1\)

Vậy phương trình có 1 nghiệm thuộc khoảng \(\left(\frac{\pi}{2};\pi\right)\)

19. \(\sin3x=\sin x\Leftrightarrow3x=\pi-x+k2\pi\Rightarrow x=\frac{\pi}{4}+\frac{1}{2}k\pi\)

33. \(DKXD:\left\{{}\begin{matrix}\cos3x\ne0\\\sin2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne\frac{\pi}{2}+k\pi\\2x\ne\pi+k\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{6}+\frac{1}{3}k\pi\\x\ne\frac{\pi}{2}+\frac{1}{2}k\pi\end{matrix}\right.\)

\(\tan3x.\cot2x=1\Leftrightarrow\tan3x=\frac{1}{\cot2x}=\tan2x\)

\(\Leftrightarrow3x=\pi+2x+k\pi\Leftrightarrow x=\pi+k\pi\) (t/m)

\(-1\le sin\left(2x+\frac{\pi}{4}\right)\le1\Rightarrow0\le y\le2\)

\(y_{min}=0\) khi \(sin\left(2x+\frac{\pi}{4}\right)=1\)

\(y_{max}=2\) khi \(sin\left(2x+\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow3\left(1-sin^22x\right)-sin2x+1=0\)

\(\Leftrightarrow-3sin^22x-sin2x+4=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{4}{3}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{4}+k\pi\)

28.

\(tan\left(2x-15^0\right)=1\Leftrightarrow2x-15^0=45^0+k180^0\)

\(\Leftrightarrow2x=60^0+k180^0\)

\(\Leftrightarrow x=30^0+k90^0\)

\(-90^0\le30^0+k90^0\le90^0\Rightarrow k=\left\{-1;0\right\}\)

\(\Rightarrow x=\left\{-60^0;30^0\right\}\Rightarrow\sum x=-30^0\)

34.

\(tan\left(x+\frac{\pi}{2}\right)=1\Leftrightarrow x+\frac{\pi}{2}=\frac{\pi}{4}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)=sin\left[2\left(-\frac{\pi}{4}+k\pi\right)-\frac{\pi}{6}\right]\)

\(=sin\left(-\frac{2\pi}{3}+k2\pi\right)=sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}\)

ĐKXĐ: ...

\(\Leftrightarrow tan\left(3x+\frac{\pi}{2}\right)=\frac{1}{cot\left(5x+1\right)}=tan\left(5x+1\right)\)

\(\Leftrightarrow5x+1=3x+\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow2x=\frac{\pi}{2}-1+k\pi\)

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

24.

Vẽ đường tròn lượng giác, kẻ đường thẳng \(cosx=\frac{13}{14}\) ta thấy cắt cung đã cho tại 3 điểm nên pt có 3 nghiệm

25.

\(sin2x=cosx=sin\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=x-\frac{\pi}{2}+n2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+n2\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}0\le\frac{\pi}{6}+\frac{k2\pi}{3}\le2\pi\\0\le-\frac{\pi}{2}+n2\pi\le2\pi\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=\left\{0;1;2\right\}\Rightarrow x=\left\{\frac{\pi}{6};\frac{5\pi}{6};\frac{3\pi}{2}\right\}\\n=1\Rightarrow x=\frac{3\pi}{2}\end{matrix}\right.\)

\(\Rightarrow T=4\pi\)

\(\Leftrightarrow sin^22x+cos^2x=sin^2x+cos^2x\)

\(\Leftrightarrow sin^22x=sin^2x\)

\(\Leftrightarrow4sin^2x.cos^2x=sin^2x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=0\\4cos^2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{3}+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Đề là chúng minh ak bạn

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=-\frac{\pi}{3}+k2\pi\\2x+\frac{\pi}{6}=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)

Do \(-\pi< x< \frac{\pi}{2}\Rightarrow x=\left\{-\frac{\pi}{4};-\frac{5\pi}{12}\right\}\)

21.

\(\Leftrightarrow\left[{}\begin{matrix}sinx+1=0\\sinx-\sqrt{2}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}>1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)

\(x\in\left[-2017;2017\right]\Rightarrow-2017\le-\frac{\pi}{2}+k2\pi\le2017\)

\(\Rightarrow\frac{\frac{\pi}{2}-2017}{2\pi}\le k\le\frac{\frac{\pi}{2}+2017}{2\pi}\)

\(\Rightarrow-320\le k\le321\) \(\Rightarrow\) pt có 642 nghiệm

22.

\(sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)

\(\Rightarrow\) Nghiệm âm lớn nhất \(x=-\frac{13\pi}{36}\) ; nghiệm dương nhỏ nhất \(x=\frac{7\pi}{36}\)

Tổng 2 nghiệm: \(-\frac{13\pi}{36}+\frac{7\pi}{36}=-\frac{\pi}{6}\)

c/

\(\Leftrightarrow1+2cos^2x-1+cosx=0\)

\(\Leftrightarrow2cos^2x-cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

d/

Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\ge0\\cosx=b\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}a+3b=2\\a^2+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2-3b\\a^2+b^2=1\end{matrix}\right.\)

\(\Rightarrow\left(2-3b\right)^2+b^2-1=0\)

\(\Rightarrow10b^2-12b+3=0\Rightarrow\left[{}\begin{matrix}b=\frac{6+\sqrt{6}}{10}\Rightarrow a=\frac{2-3\sqrt{6}}{10}\left(l\right)\\b=\frac{6-\sqrt{6}}{10}\Rightarrow a=\frac{2+3\sqrt{6}}{10}\end{matrix}\right.\)

\(\Rightarrow cosx=\frac{6-\sqrt{6}}{10}\)

\(\Rightarrow x=\pm arccos\left(\frac{6-\sqrt{6}}{10}\right)+k2\pi\)

b/

\(cos\left(8sinx\right)=1\)

\(\Leftrightarrow8sinx=k2\pi\)

\(\Leftrightarrow sinx=\frac{k\pi}{4}\)

Do \(-1\le sinx\le1\Rightarrow-1\le\frac{k\pi}{4}\le1\)

\(\Rightarrow k=\left\{-1;0;1\right\}\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{\pi}{4}\\sinx=0\\sinx=\frac{\pi}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=\pi\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=k\pi\end{matrix}\right.\)

14) đề bài tương đương với

cotx = -\(\sqrt{3}\Leftrightarrow cotx=cot\left(\frac{-\pi}{6}\right)\Leftrightarrow x=\frac{-\pi}{6}+k\pi\)

16) đề bài tương đương với

cosx=2m-1

ta có cosx \(\in\left[-1;1\right]\Leftrightarrow-1\le2m-1\le1\\ \Leftrightarrow\left\{{}\begin{matrix}2m-1\ge-1\\2m-1\le1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\m\le1\end{matrix}\right.\)

vì n là số nguyên nên m=0 hoặc m=1