1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

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

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

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

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

1/ \(sinx=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\)

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

b/ \(cos=-\frac{\sqrt{2}}{2}=cos\left(\frac{3\pi}{4}\right)\)

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

c/ \(tanx=\sqrt{3}=tan\left(\frac{\pi}{3}\right)\)

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

d/ \(cotx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)

2/

a/ \(sin^2x+sinx-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-2\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{2}+k2\pi\)

b/ \(cot^2x-2cotx-3=0\)

\(\Leftrightarrow\left(cotx+1\right)\left(cotx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot3+k\pi\end{matrix}\right.\)

3/ \(\Leftrightarrow1-cos2x+1-cos4x+1-cos6x=3\)

\(\Leftrightarrow cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2coss4x.cos2x+cos4x=0\)

\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\frac{2\pi}{3}+k2\pi\\2x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

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

c.

\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)

\(\Leftrightarrow2cos\left(x+12^0\right)=1\)

\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)

2.

Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:

\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)

\(\Rightarrow-1\le m\le\dfrac{1}{2}\)

a.

\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

b.

\(2x-10^0=arccot\left(4\right)+k180^0\)

\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)

2.

Phương trình \(sin\left(3x-27^o\right)=2m^2+m\) có nghiệm khi:

\(2m^2+m\in\left[-1;1\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m\le1\\2m^2+m\ge-1\end{matrix}\right.\)

\(\Leftrightarrow\left(m+1\right)\left(2m-1\right)\le0\)

\(\Leftrightarrow-1\le m\le\dfrac{1}{2}\)

a) \(\sin 2x + 1 - 2{\sin ^2}2x = 0\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin 2x = 1}\\{\sin 2x =  - \frac{1}{2}}\end{array}\;\;\;} \right. \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{\sin 2x = \sin \frac{\pi }{2}}\\{\sin 2x = \sin  - \frac{\pi }{6}}\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = \frac{\pi }{2} + k2\pi }\\{2x =  - \frac{\pi }{6} + k2\pi }\\{2x = \pi  + \frac{\pi }{6} + k2\pi }\end{array}} \right.\;\;\)

\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{4} + k2\pi }\\{x =  - \frac{\pi }{{12}} + k\pi }\\{x = \frac{{7\pi }}{{12}} + k\pi }\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

b) \(\cos 3x =  - \cos 7x\; \Leftrightarrow \cos 3x + \cos 7x = 0\;\; \Leftrightarrow 2\cos 5x\cos 2x = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos 5x = 0}\\{\cos 2x = 0\;}\end{array}} \right.\;\;\)

\( \Leftrightarrow \left[ \begin{array}{l}\cos 5x = \cos \frac{\pi }{2}\\\cos 2x = \cos \frac{\pi }{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\5x =  - \frac{\pi }{2} + k2\pi \\2x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x =  - \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = \frac{\pi }{4} + k\pi \\x =  - \frac{\pi }{4} + k\pi \end{array} \right.;k \in Z\)

a/

\(\Leftrightarrow3\left(cos4x+1\right)+2cos^2x\left(1-4cos^4x\right)=0\)

\(\Leftrightarrow3\left(2cos^22x-1+1\right)+2cos^2x\left(1-2cos^2x\right)\left(1+2cos^2x\right)=0\)

\(\Leftrightarrow6cos^22x+\left(1+cos2x\right).\left(-cos2x\right)\left(2+cos2x\right)=0\)

Đặt \(cos2x=a\)

\(\Rightarrow6a^2-a\left(a+1\right)\left(a+2\right)=0\)

\(\Leftrightarrow a\left(-a^2+3a-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\\cos2x=2\left(l\right)\end{matrix}\right.\)

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

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

b/

\(\Leftrightarrow4+3sinx+sin^3x=3\left(1-sin^2x\right)+\left(1-sin^2x\right)^3\)

Đặt \(sinx=a\) ta được:

\(a^3+3a+4=3-3a^2+\left(1-a\right)^3\)

\(\Leftrightarrow a^3+3a^2+3a+1=\left(1-a\right)^3\)

\(\Leftrightarrow\left(a+1\right)^3=\left(1-a\right)^3\)

\(\Leftrightarrow a+1=1-a\)

\(\Leftrightarrow a=0\)

\(\Rightarrow sinx=0\Rightarrow x=k\pi\)

c/

ĐKXĐ: ...

\(\Leftrightarrow2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=2cos^2x-4\)

\(\Leftrightarrow2cos^2x+2cos^2x.tanx.tan\frac{x}{2}=2cos^2x-4\)

\(\Leftrightarrow cos^2x.tanx.tan\frac{x}{2}=-2\)

\(\Leftrightarrow sinx.cosx.tan\frac{x}{2}=-2\)

\(\Leftrightarrow sinx.cosx.\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=-2\)

\(\Leftrightarrow sinx.cosx.\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}.cos\frac{x}{2}}=-1\)

\(\Leftrightarrow cosx\left(\frac{1-cosx}{2}\right)=-1\)

\(\Leftrightarrow cos^2x-cosx-2=0\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pi+k2\pi\)