So sánh :
a) \(\left(0,01\right)^{-\sqrt{3}}\) và 1000
b) \(\left(\frac{\pi}{2}\right)^{2\sqrt{2}}\) và \(\left(\frac{\pi}{2}\right)^3\)
Tính:
a) \(\cos \left( {a + \frac{\pi }{6}} \right)\), biết \(\sin a = \frac{1}{{\sqrt 3 }}\) và \(\frac{\pi }{2} < a < \pi \);
b) \(\tan \left( {a - \frac{\pi }{4}} \right)\), biết \(\cos a = - \frac{1}{3}\) và \(\pi < a < \frac{{3\pi }}{2}\).
a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\). Do đó \(\cos a = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \frac{1}{3}} = - \frac{{\sqrt 6 }}{3}\)
Ta có: \(\cos \left( {a + \frac{\pi }{6}} \right) = \cos a\cos \frac{\pi }{6} - \sin a\sin \frac{\pi }{6} = - \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 3 }}.\frac{1}{2} = - \frac{{\sqrt 3 + 3\sqrt 2 }}{6}\)
b) Vì \(\pi < a < \frac{{3\pi }}{2}\) nên \(\sin a < 0\). Do đó \(\sin a = \sqrt {1 - {{\cos }^2}a} = \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3}\)
Suy ra \(\tan a\; = \frac{{\sin a}}{{\cos a}} = \frac{{ - \frac{{2\sqrt 2 }}{3}}}{{ - \frac{1}{3}}} = 2\sqrt 2 \)
Ta có: \(\tan \left( {a - \frac{\pi }{4}} \right) = \frac{{\tan a - \tan \frac{\pi }{4}}}{{1 + \tan a\tan \frac{\pi }{4}}} = \frac{{\frac{{\sin a}}{{\cos a}} - 1}}{{1 + \frac{{\sin a}}{{\cos a}}}} = \frac{{2\sqrt 2 - 1}}{{1 + 2\sqrt 2 }} = \frac{{9 - 4\sqrt 2 }}{7}\)
Giải phương trình:
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
d) \(2\cos 3x + 5 = 3\)
e) \(3\tan x = - \sqrt 3 \)
g) \(\cot x - 3 = \sqrt 3 \left( {1 - \cot x} \right)\)
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
d) \(2\cos 3x + 5 = 3\)
\(\begin{array}{l} \Leftrightarrow \cos 3x = - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
e)
\(\begin{array}{l}3\tan x = - \sqrt 3 \\ \Leftrightarrow \tan x = \frac{{ - \sqrt 3 }}{3}\\ \Leftrightarrow \tan x = \tan \left( { - \frac{\pi }{6}} \right)\\ \Leftrightarrow x = - \frac{\pi }{6} + k\pi \end{array}\)
g)
\(\begin{array}{l}\cot x - 3 = \sqrt 3 \left( {1 - \cot x} \right)\\ \Leftrightarrow \cot x - 3 = \sqrt 3 - \sqrt 3 \cot x\\ \Leftrightarrow \cot x + \sqrt 3 \cot x = \sqrt 3 + 3\\ \Leftrightarrow (1 + \sqrt 3 )\cot x = \sqrt 3 + 3\\ \Leftrightarrow \cot x = \sqrt 3 \\ \Leftrightarrow \cot x = \cot \frac{\pi }{6}\\ \Leftrightarrow x = \frac{\pi }{6} + k\pi \end{array}\)
\(2\sin^2\left(5\pi+1\right)-\left(\sqrt{3}+1\right)\sin2\left(\frac{\pi}{2}-x\right)+\sqrt{3}\sin^2\left(\frac{9\pi}{2}+x\right)=0\)
Đề đúng là \(2sin^2\left(5\pi+1\right)\) chứ bạn?
Chứ thấy nó hơi thế nào ấy?
So sánh các cặp số :
a) \(\sqrt[3]{10}\) và \(\sqrt[5]{20}\)
b) \(\left(\frac{1}{e}\right)^{\sqrt{8}-3}\) và 1
c) \(\left(\frac{1}{8}\right)^{\pi}\) và \(\left(\frac{1}{8}\right)^{3.14}\)
d) \(\left(\frac{1}{\pi}\right)^{1.4}\) và \(\pi^{-\sqrt{2}}\)
a) \(\sqrt[3]{10}=\sqrt[15]{10^5}>\sqrt[15]{20^3=\sqrt[5]{20}}\)
b) Vì \(\frac{1}{e}<1\) và \(\sqrt{8}-3<0\) nên \(\left(\frac{1}{e}\right)^{\sqrt{8}-3}>1\)
c) Vì \(\frac{1}{8}<1\) và \(\pi>3.14\) nên \(\left(\frac{1}{8}\right)^{\pi}<\left(\frac{1}{8}\right)^{3,14}\)
d) Vì \(\frac{1}{\pi}<1\) và \(1,4<\sqrt{2}\) nên \(\left(\frac{1}{\pi}\right)^{1,4}>\pi^{-\sqrt{2}}\)
1) \(sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right).tan^2x-cos^2\frac{x}{2}=0\)
2) \(tanx=sin^2x\left(c-\frac{\pi}{2010}\right)+cos^2\left(2x+\frac{\pi}{2010}\right)+sinx.sin\left(3x+\frac{\pi}{1005}\right)\)
3) \(1+2cosx\left(sinx-1\right)+\sqrt{2}sinx+4cosx.sin^2\frac{x}{2}=0\)
4) \(3cos4x-8cos^6x+2cos4x=3\)
5) \(1+sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)\)
6) \(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-4\sqrt{3}cos^2x.sinx.cos2x\)
7) \(\frac{tan^2x+tanx}{tan^2x+1}=\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{4}\right)\)
8) \(cos^4x+sin^4x+cos\left(x-\frac{\pi}{4}\right).sin\left(3x-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
6.
\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)
\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)
\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
giải các pt
a) \(cos^2\left(\frac{\pi}{3}+x\right)+4cos\left(\frac{\pi}{6}-x\right)=4\)
b) \(5cos\left(2x+\frac{\pi}{3}\right)=4sin\left(\frac{5\pi}{6}-x\right)-9\)
c) \(2sin^2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
d) \(2sin^2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
a/
Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)
Pt trở thành:
\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)
\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)
\(\Leftrightarrow cos^2a+4sina-4=0\)
\(\Leftrightarrow1-sin^2a+4sina-4=0\)
\(\Leftrightarrow-sin^2a+4sina-3=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=1\)
\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)
b/
Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)
Pt trở thành:
\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)
\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)
\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)
\(\Leftrightarrow10sin^2a+4sina-14=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)
\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
Mạch điện gồm điện trở \(R=30\sqrt{3}\Omega\) nối tiếp với tụ điện có \(C=\frac{10^{-3}}{3\pi}F\). Điện áp tức thời ở hai đầu đoạn mạch là \(u=120\sqrt{2}\cos100\pi t\left(V\right)\) .Dòng điện qua mạch có biểu thức:
A. \(i=2\sqrt{2}\cos\left(100\pi t+\frac{\pi}{6}\right)\left(A\right)\)
B. \(i=2\sqrt{2}\cos\left(100\pi t+\frac{\pi}{3}\right)\left(A\right)\)
C. \(i=3\sqrt{2}\cos\left(100\pi t+\frac{\pi}{6}\right)\left(A\right)\)
D. \(i=3\sqrt{2}\cos\left(100\pi t+\frac{\pi}{3}\right)\left(A\right)\)
Ta có: \(Z_C=\frac{1}{C\omega}=30\Omega\)
\(\tan\varphi=-\frac{Z_c}{R}=-\frac{1}{\sqrt{3}}\)
\(\Rightarrow\varphi=-\frac{\pi}{6}\)
\(\Rightarrow\varphi_U-\varphi_I=-\frac{\pi}{6}\Rightarrow\varphi_1=\frac{\pi}{6}rad\)
Lại có: \(I=\frac{U}{Z}=2\sqrt{2}\left(A\right)\)
\(\Rightarrow i=2\sqrt{2}\cos\left(100\pi t+\frac{\pi}{6}\right)\left(A\right)\)
Đáp án A
Giair các pt lượng giác sau:
1) \(sin\left(x-\frac{\pi}{4}\right)\left(2cos+\sqrt{2}\right)tan2x=0\)
2) \(tan2x.sinx+3\left(sin-\sqrt{3}tan2x\right)-3\sqrt{3}=0\)
3) \(\frac{cos2x}{sin\left(x+\frac{3\pi}{4}\right)}=\frac{sin\left(x+\frac{3\pi}{4}\right)}{cos2x}\)
4) \(\left(\frac{tanx-1}{tanx+1}+cot2x\right)\left(3tan-\sqrt{3}\right)=0;0< x< \pi\)
a/ ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\2cosx+\sqrt{2}=0\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\cosx=cos\left(\frac{3\pi}{4}\right)\\2x=k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=\frac{3\pi}{4}+k2\pi\left(l\right)\\x=-\frac{3\pi}{4}+k2\pi\left(l\right)\\x=\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x=\frac{k\pi}{2}\)
b/
ĐKXĐ: \(x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\Leftrightarrow tan2x.sinx+3sinx-\sqrt{3}tan2x-3\sqrt{3}=0\)
\(\Leftrightarrow sinx\left(tan2x+3\right)-\sqrt{3}\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\right)\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\sqrt{3}>1\left(vn\right)\\tan2x=-3\end{matrix}\right.\)
\(\Rightarrow2x=arctan\left(-3\right)+k\pi\)
\(\Rightarrow x=\frac{arctan\left(-2\right)}{2}+\frac{k\pi}{2}\)
c/
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x+\frac{3\pi}{4}\right)\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\frac{3\pi}{4}\ne k\pi\\2x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne-\frac{3\pi}{4}+k\pi\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(cos^22x=sin^2\left(x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{3\pi}{2}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{3\pi}{2}\right)=cos\left(2x+\frac{\pi}{2}\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=2x+\frac{\pi}{2}+k2\pi\\4x=-2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)
1) \(4cos^24x+2\left(\sqrt{3}+\sqrt{2}\right)cos4x+\sqrt{6}=0\)
2) \(cos4x+2+sin\left(2x+\frac{3\pi}{2}\right)=2cos^2x\)
3) \(sin\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin\left(\frac{\pi}{6}-x\right)=1\)
4) \(2cos\left(4x-\frac{\pi}{3}\right)+4cos2x=-1\)
5) \(cos^22x+cos^23x=sin^2x\)
6) \(sinx+\left(\sqrt{2}-1\right)cosx=1\)
7) \(cos2x-\left(\sqrt{3}+1\right)cosx+\frac{2+\sqrt{3}}{2}=0\)
1.
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)
2.
\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)
\(\Leftrightarrow2cos^22x-cos2x=cos2x\)
\(\Leftrightarrow cos^22x-cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
3.
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow...\)
4.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
5.
\(cos^22x+\frac{1}{2}+\frac{1}{2}cos6x=\frac{1}{2}-\frac{1}{2}cos2x\)
\(\Leftrightarrow cos^22x+\frac{1}{2}\left(cos6x+cos2x\right)=0\)
\(\Leftrightarrow cos^22x+cos4x.cos2x=0\)
\(\Leftrightarrow cos2x\left(cos2x+cos4x\right)=0\)
\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)