\(a,3^{x+2}=7\\ \Leftrightarrow x+2=log_37\\ \Leftrightarrow x=log_37-2\approx-0.229\)

\(b,3\cdot10^{2x+1}=5\\ \Leftrightarrow10^{2x+1}=\dfrac{5}{3}\\ \Leftrightarrow2x+1=log\left(\dfrac{5}{3}\right)\\ \Leftrightarrow2x=log\left(\dfrac{5}{3}\right)-1\\ \Leftrightarrow x=\dfrac{1}{2}\cdot log\dfrac{5}{3}-\dfrac{1}{2}\\ \Leftrightarrow x\approx-0,389\)

a) \(\cos \left( {3x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} =  - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi  + k2\pi }\\{3x =  - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)

\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x =  - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} =  - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} =  - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x =  - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi  + k2\pi }\\{x =  - \pi  + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)

c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x =  - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)

a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi  - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)

b) \(2\cos x =  - \sqrt 2 \;\; \Leftrightarrow \cos x =  - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x =  - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)

c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)

\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)

\(a,sin2x-2sinx+cosx-1=0\)

\(\Leftrightarrow2sinxcosx-2sinx+cosx-1=0\)

\(\Leftrightarrow2sinx\left(cosx-1\right)+cosx-1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}cosx=1\\sinx=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2k\pi\\x=\frac{-\pi}{6}+2k\pi\end{cases}}}\)

\(b,\sqrt{2}\left(sinx-2cosx\right)=2-sin2x\)

\(\Leftrightarrow\sqrt{2}sinx-2\sqrt{2}cosx-2+2sinxcosx=0\)

\(\Leftrightarrow\sqrt{2}sinx\left(1+\sqrt{2}cosx\right)-2.\left(\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{2}cosx+1\right)\left(\sqrt{2}sinx-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{-\sqrt{2}}{2}\\sinx=\frac{2\sqrt{2}}{2}\left(l\right)\end{cases}}\)(vì \(-1\le sinx\le1\))

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3\pi}{4}+2k\pi\\x=\frac{5\pi}{4}+2k\pi\end{cases}}\)

\(c,\frac{1}{cosx}-\frac{1}{sinx}=2\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

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

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

\(\Leftrightarrow sin2x+1=0\)

\(\Leftrightarrow sin2x=-1\)

\(\Leftrightarrow2x=\frac{3\pi}{2}+2k\pi\)

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

a.

\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

\(\Leftrightarrow cos^2x=0\)

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

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

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

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

a) Trong Hình 5, M là điểm biểu diễn của góc lượng giác \(\alpha \) trên đường tròn lượng giác. Ta có:

OK = MH = \(\sin \alpha \)

OH = KM = \(\cos \alpha \)

\(\begin{array}{l}O{M^2} = O{H^2} + M{H^2}\\ \Rightarrow 1 = {\sin ^2}\alpha  + {\cos ^2}\alpha \end{array}\)

b) \(1 + {\tan ^2}\alpha  = \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{1}{{{{\cos }^2}\alpha }}\)