Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

Đk: \(x\ne\dfrac{\pi}{2}+k\pi\) \(,k\in Z\)

Pt \(\Leftrightarrow\dfrac{sinx}{cosx}-\sqrt{3}=\dfrac{1}{cosx}\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\),\(k\in Z\)  kết hợp với đk \(\Rightarrow x=\dfrac{7\pi}{6}+k2\pi\), \(k\in Z\)

Vậy...

53.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(2tanx-\dfrac{2}{tanx}-3=0\)

\(\Leftrightarrow2tan^2x-3tanx-2=0\)

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

Trong khoảng \(\left(-\dfrac{\pi}{2};\pi\right)\) , pt \(tanx=k\) ứng với mỗi \(k>0\) sẽ có 1 nghiệm (nằm wor góc phần tư thứ nhất) và \(k< 0\) có 2 nghiệm (nằm ở góc phần tư thứ 2 và 4)

\(\Rightarrow\) Pt đã cho có 3 nghiệm

54, Đk: \(x\ne\pm\dfrac{\pi}{4}+k\pi\)\(;x\ne\dfrac{\pi}{2}+k\pi\)

Pt \(\Leftrightarrow\dfrac{tan2x}{2}=\dfrac{1}{2}.cot\left[\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}-x\right)\right]\)

\(\Leftrightarrow tan2x=tan\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\)

Ý D (bài này nếu vào tự luận thì không bt có phải loại đk ko?)

Đặt \(x+\dfrac{\pi}{3}=t\Rightarrow\dfrac{\pi}{6}-x=\dfrac{\pi}{2}-t\)

Ta được:

\(cos2t+4cos\left(\dfrac{\pi}{2}-t\right)=\dfrac{5}{2}\)

\(\Leftrightarrow1-2sin^2t+4sint=\dfrac{5}{2}\)

\(\Leftrightarrow4sin^2t-8sint+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sint=\dfrac{3}{2}>1\left(ktm\right)\\sint=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

C48)

Tại \(x\in\left(-\dfrac{\pi}{2};0\right)\Rightarrow sinx\in\left(-1;0\right)\)

Đặt \(t=sinx\) \(;t\in\left(-1;0\right)\)

Pttt: \(2t^2-\left(2m+1\right)t+m=0\) (*)

Để pt ban đầu có nghiệm \(\in\left(-\dfrac{\pi}{2};0\right)\) khi pt (*) có nghiệm t thỏa mãn \(-1< t< 0\)

Có \(\Delta=\left(2m-1\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{2m+1-\left(2m-1\right)}{4}=\dfrac{1}{2}\\t_2=\dfrac{2m+1+\left(2m-1\right)}{4}=m\end{matrix}\right.\) 

Vì \(t_1=\dfrac{1}{2}\notin\left(-1;0\right)\)

Để pt có nghiệm thỏa mãn đk\(\Leftrightarrow m\in\left(-1;0\right)\)

Ý A

1.

Chắc đề là \(sin\left[\pi sin2x\right]=1?\)

\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)

Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)

\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)

Thế vào (1)

\(\Rightarrow sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)

Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)

Thế vào (2):

\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\) chắc bạn tự giải tiếp được

3.

\(\Leftrightarrow2sin\left(x+84^0\right).cos\left(60^0\right)=cos20^0\)

\(\Leftrightarrow sin\left(x+84^0\right)=sin70^0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+84^0=70^0+k360^0\\x+84^0=110^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-14^0+k360^0\\x=26^0+k360^0\end{matrix}\right.\)

1.

\(\Leftrightarrow2sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)

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

\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))

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

2.

\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)

Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)

3. Làm tương tự câu 2

4.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Leftrightarrow\left(\dfrac{1+cos6x}{2}\right)cos2x-\dfrac{1+cos2x}{2}=0\)

\(\Leftrightarrow cos2x+cos2x.cos6x-cos2x-1=0\)

\(\Leftrightarrow\dfrac{1}{2}cos8x+\dfrac{1}{2}cos4x-1=0\)

\(\Leftrightarrow cos8x+cos4x-2=0\)

\(\Leftrightarrow2cos^24x-1+cos4x-2=0\)

\(\Leftrightarrow2cos^24x+cos4x-3=0\)

\(\Rightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\dfrac{3}{2}< -1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow4x=k2\pi\)

\(\Rightarrow x=\dfrac{k\pi}{2}\)

\(cos3x+sin3x=4cos^3x-3cosx+3sinx-4sin^3x\\ =4\left(cosx-sinx\right)\left(1+sinxcosx\right)-3\left(cosx-sinx\right)\\ =\left(cosx-sinx\right)\left(1+4sinxcosx\right)\)

pt thành: 

\(5sinx+5cosx-5sinx=cos2x+3\\ \Leftrightarrow2cos^2x-5cosx+2=0\\ \Leftrightarrow cosx=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi,k\in Z\)

+xét \(0< \dfrac{\pi}{3}+k2\pi< 2\pi\\ \Leftrightarrow-\dfrac{1}{6}< k< \dfrac{5}{6}\Leftrightarrow k=0\Rightarrow x=\dfrac{\pi}{3}\)

+xet \(0< -\dfrac{\pi}{3}+k2\pi< 2\pi\Leftrightarrow\dfrac{1}{6}< k< \dfrac{7}{6}\Leftrightarrow k=1\\ \Rightarrow x=\dfrac{5\pi}{3}\)

vay..

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(\dfrac{4sin^22x+6sin^2x-9-3cos2x}{cosx}=0\)

\(\Leftrightarrow4sin^22x+6sin^2x-9-3cos2x=0\)

\(\Leftrightarrow4\left(sin^22x-1\right)+3\left(2sin^2x-1\right)-3cos2x-2=0\)

\(\Leftrightarrow-4cos^22x-3cos2x-3cos2x-2=0\)

\(\Leftrightarrow2cos^22x+3cos2x+1=0\)

\(\Leftrightarrow\left(2cos2x+1\right)\left(cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-\dfrac{1}{2}\\cos2x=-1\end{matrix}\right.\)

Với \(cos2x=-\dfrac{1}{2}\Leftrightarrow2x=\pm\dfrac{2\pi}{3}+k2\pi\Leftrightarrow x=\pm\dfrac{\pi}{3}+k\pi\left(tm\right)\)

Với \(cos2x=-1\Leftrightarrow2x=\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(tm\right)\)

Vậy ...

ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3}\left(1+cot^2x\right)=3cotx+\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}cot^2x-3cotx=0\)

\(\Leftrightarrow\sqrt{3}cotx\left(cotx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=0\\cotx=\sqrt{3}\end{matrix}\right.\)

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

\(2sin^2x-3sinx+1=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))  (I)

Có \(0\le x< \dfrac{\pi}{2}\)\(\Leftrightarrow\left[{}\begin{matrix}0\le\dfrac{\pi}{2}+k2\pi< \dfrac{\pi}{2}\\0\le\dfrac{\pi}{6}+k2\pi< \dfrac{\pi}{2}\\0\le\dfrac{5\pi}{6}+k2\pi< \dfrac{\pi}{2}\end{matrix}\right.\)(\(k\in Z\))

\(\Leftrightarrow\left[{}\begin{matrix}0\le\dfrac{1}{2}+2k< \dfrac{1}{2}\\0\le\dfrac{1}{6}+2k< \dfrac{1}{2}\\0\le\dfrac{5}{6}+2k< \dfrac{1}{2}\end{matrix}\right.\) (\(k\in Z\)) \(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{4}\le k< 0\left(1\right)\\-\dfrac{1}{12}\le k< \dfrac{1}{6}\left(2\right)\\-\dfrac{5}{12}\le k< -\dfrac{1}{6}\left(3\right)\end{matrix}\right.\)(\(k\in Z\))

Do k nguyên, từ (1) và (3) \(\Rightarrow k\in\varnothing\)

Từ (2)\(\Rightarrow k=0\)\(\Rightarrow x=\dfrac{\pi}{6}+0.2\pi=\dfrac{\pi}{6}\)

Ý C 

(Hoặc sau khi bạn làm đến đoạn số (I),bạn vẽ đường tròn lượng giác ra sẽ tìm được x)