1.

\(\Leftrightarrow1-cos^22x-2\left(\frac{1+cos2x}{2}\right)+\frac{3}{4}=0\)

\(\Leftrightarrow-cos^22x-cos2x+\frac{3}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{1}{2}\\cos2x=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=\pm\frac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\frac{\pi}{6}+k\pi\)

2.

\(2\left(2cos^2x-1\right)+2cosx-\sqrt{2}=0\)

\(\Leftrightarrow4cos^2x+2cosx-2-\sqrt{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{2}}{2}\\cosx=-\frac{1+\sqrt{2}}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=-\frac{\pi}{4}+l2\pi\end{matrix}\right.\) mà \(-\frac{\pi}{2}< x< \frac{5\pi}{2}\Rightarrow\left\{{}\begin{matrix}-\frac{\pi}{2}< \frac{\pi}{4}+k2\pi< \frac{5\pi}{2}\\-\frac{\pi}{2}< -\frac{\pi}{4}+l2\pi< \frac{5\pi}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0;1\\l=0;1\end{matrix}\right.\) \(\Rightarrow x=\left\{\frac{\pi}{4};\frac{9\pi}{4};-\frac{\pi}{4};\frac{7\pi}{4}\right\}\)

Có 4 nghiệm

3. ĐKXĐ: ...

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-\frac{1}{2}\right)+k\pi\\x=arctan\left(2\right)+k\pi\end{matrix}\right.\)

Có 3 nghiệm trong khoảng đã cho \(x=arctan\left(-\frac{1}{2}\right);x=arctan\left(-\frac{1}{2}\right)+\pi;x=arctan\left(2\right)\)

4. ĐKXĐ: ...

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

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

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

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

Nghiệm âm lớn nhất của pt là \(x=-\frac{\pi}{2}\)

5. ĐKXĐ; ...

\(\Leftrightarrow tan^2x-\left(1+\sqrt{3}\right)tanx+\sqrt{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+l\pi\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-2019\pi< \frac{\pi}{4}+k\pi< 2019\pi\\-2019\pi< \frac{\pi}{3}+l\pi< 2019\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-2019\le k\le2018\\-2019\le l\le2018\end{matrix}\right.\)

Tổng các nghiệm: \(2.\left(-2019\pi\right)+4038\left(\frac{\pi}{3}+\frac{\pi}{4}\right)=-\frac{3365\pi}{2}< -3\)

Đáp án A đúng

98:

$-2\cos 2x=1$

$\Leftrightarrow \cos 2x=\frac{-1}{2}$

\(\Rightarrow \left[\begin{matrix} 2x=\frac{2}{3}\pi +2k\pi\\ 2x=\frac{-2}{3}\pi +2k\pi\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{\pi}{3}+k\pi\\ x=\frac{-\pi}{3}+k\pi \end{matrix}\right.\) với $k$ nguyên.

97:

$\cos x=0\Rightarrow x=\pi (k+\frac{1}{2})$ với $k\in \mathbb{Z}$

Vì $x\in (0;2018\pi)$ nên $0< \pi (k+\frac{1}{2})< 2018\pi$

$\Leftrightarrow 0< k+\frac{1}{2}< 2018$

$\Leftrightarrow \frac{-1}{2}< k< \frac{4035}{2}$
Vì $k$ nguyên nên $k\in\left\{0;1;2;...; 2017\right\}$

Có 2018 giá trị của $k$ thỏa mãn kéo theo có 2018 giá trị $x$ thỏa mãn điều kiện đề bài.

96:

$\sin x=1\Rightarrow x=\pi (\frac{1}{2}+2k)$ với $k$ là số nguyên.

$x\in (0;2018\pi)$

$\Leftrightarrow \frac{1}{2}+2k\in (0;2018)$

$\Leftrightarrow k\in (-\frac{1}{4};\frac{4035}{4})$

Vì $k$ nguyên nên $k\in\left\{0;1;...;1008\right\}$

Như vậy có 1009 giá trị $k$ thỏa mãn kéo theo có 1008 giá trị $x$

Vậy có 1009 nghiệm của PT $\sin x=1$ thuộc khoảng $(0;2018\pi)$

1.

Đề là \(x\in\left(0;\frac{\pi}{4}\right)\) hay \(x\in\left[0;\frac{\pi}{4}\right]\) ?

2.

\(sin3x-4sinx.cos2x=0\)

\(\Leftrightarrow sin3x-\left(2sin3x-2sinx\right)=0\)

\(\Leftrightarrow2sinx-sin3x=0\)

\(\Leftrightarrow2sinx-3sinx+4sin^3x=0\)

\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)

\(\Leftrightarrow sinx\left(1-2cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

3.

\(sin^2x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

4.

\(\sqrt{3}sin2x+1-cos2x=3\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

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

5.

Ko có 4 đáp án thì làm sao biết, có vô số pt tương đương với pt này :)

6.

\(sinx+cosx-2sinx.cosx+1=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t+1-t^2+1=0\)

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

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

\(\Rightarrow2sinx.cosx=t^2-1=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

107.

\(\Leftrightarrow tan2x=-\sqrt{3}\)

\(\Leftrightarrow2x=-\frac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=-\frac{\pi}{6}+\frac{k\pi}{2}\)

\(2000\pi\le-\frac{\pi}{6}+\frac{k\pi}{2}\le2018\pi\)

\(\Leftrightarrow4000+\frac{1}{3}\le k\le4036+\frac{1}{2}\)

Có \(4036-4001+1=36\) nghiệm

108.

\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{4}+k2\pi\\5x=-\frac{\pi}{4}+n2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{20}+\frac{k2\pi}{5}\\x=-\frac{\pi}{20}+\frac{n2\pi}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-50\pi\le\frac{\pi}{20}+\frac{k2\pi}{5}\le0\\-50\pi\le-\frac{\pi}{20}+\frac{n2\pi}{5}\le0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-125-\frac{1}{8}\le k\le-\frac{1}{8}\\-125+\frac{1}{8}\le n\le\frac{1}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-125\le k\le-1\\-124\le n\le0\end{matrix}\right.\)

Có \(-1-\left(-125\right)+1+0-\left(-124\right)+1=250\) nghiệm

109.

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

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

\(\Rightarrow\left[{}\begin{matrix}0< -\frac{\pi}{12}+k\pi< \pi\\0< \frac{7\pi}{12}+k\pi< \pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{12}< k< \frac{13}{12}\\-\frac{7}{12}< k< \frac{5}{12}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}k=1\\k=0\end{matrix}\right.\) có 2 nghiệm

110.

\(\Leftrightarrow cos2x=-\frac{1}{2}\)

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

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

Ko có đáp án chọn nên ko thể bấm được, chỉ giải được tự luận thôi :)

Với \(x\in\left(-\frac{\pi}{4};\frac{\pi}{2}\right)\Rightarrow cosx>0\Rightarrow3cosx+1>0\)

Do đó pt tương đương:

\(2cos2x-1=0\Rightarrow cos2x=\frac{1}{2}\)

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

Pt có 2 nghiệm thuộc khoảng đã cho là \(x=\left\{-\frac{\pi}{6};\frac{\pi}{6}\right\}\)

a/ ĐKXĐ:...

\(\Leftrightarrow\frac{sinx}{cosx}-\frac{\sqrt{2}}{cosx}=1\)

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

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

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

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{3\pi}{4}+k2\pi\)

b/

ĐKXĐ: ...

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)

\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)

\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)

c/

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

\(\Leftrightarrow sinx=\frac{\sqrt{2}}{2}\)

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

d/

\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)

\(\Leftrightarrow sin2x+cos2x=1\)

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

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

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

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

c/

Hình như câu này đề sai

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

\(\Leftrightarrow sinx+cosx-\left(sinx+cosx\right)=\sqrt{2}\)

\(\Leftrightarrow0=\sqrt{2}\)

Pt vô nghiệm

d/ Hình như câu này đề cũng sai

\(\Leftrightarrow sin2x-2cos2x-5=0\)

\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\sqrt{5}\)

\(\Leftrightarrow sin\left(2x-a\right)=\sqrt{5}\) (với \(sina=\frac{2}{\sqrt{5}};cosa=\frac{1}{\sqrt{5}}\))

Pt vô nghiệm do \(\sqrt{5}>1\)