1.

PT $\Leftrightarrow 2^{x^2-5x+6}+2^{1-x^2}-2^{7-5x}-1=0$

$\Leftrightarrow (2^{x^2-5x+6}-2^{7-5x})-(1-2^{1-x^2})=0$

$\Leftrightarrow 2^{7-5x}(2^{x^2-1}-1)-(2^{x^2-1}-1)2^{1-x^2}=0$

$\Leftrightarrow (2^{x^2-1}-1)(2^{7-5x}-2^{1-x^2})=0$

$\Rightarrow 2^{x^2-1}-1=0$ hoặc $2^{7-5x}-2^{1-x^2}=0$

Nếu $2^{x^2-1}=1\Leftrightarrow x^2-1=0$

$\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$

$2^{7-5x}-2^{1-x^2}=0$

$\Leftrightarrow 7-5x=1-x^2\Leftrightarrow x^2-5x+6=0$

$\Leftrightarrow (x-2)(x-3)=0\Leftrightarrow x=2; x=3$

2. Đặt $\sin ^2x=a$ thì $\cos ^2x=1-a$. PT trở thành:

$16^a+16^{1-a}=10$

$\Leftrightarrow 16^a+\frac{16}{16^a}=10$

$\Leftrightarrow (16^a)^2-10.16^a+16=0$

Đặt $16^a=x$ thì:

$x^2-10x+16=0$

$\Leftrightarrow (x-2)(x-8)=0$

$\Leftrightarrow x=2$ hoặc $x=8$

$\Leftrightarrow 16^a=2$ hoặc $16^a=8$

$\Leftrightarrow 2^{4a}=2$ hoặc $2^{4a}=2^3$

$\Leftrightarroww 4a=1$ hoặc $4a=3$

$\Leftrightarrow a=\frac{1}{4}$ hoặc $a=\frac{3}{4}$

Nếu $a=\frac{1}{4}\Leftrightarrow \sin ^2x=\frac{1}{4}$

$\Leftrightarrow \sin x=\pm \frac{1}{2}$

Nếu $a=\sin ^2x=\frac{3}{4}\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$

Đến đây thì đơn giản rồi.

a) Vì \(\sin \frac{\pi }{6} = \frac{1}{2}\) nên ta có phương trình \(sin2x = \sin \frac{\pi }{6}\)

\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{6} + k2\pi \\2x = \pi  - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

\(\begin{array}{l}b,\,\,sin(x - \frac{\pi }{7}) = sin\frac{{2\pi }}{7}\\ \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{7} = \frac{{2\pi }}{7} + k2\pi \\x - \frac{\pi }{7} = \pi  - \frac{{2\pi }}{7} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{3\pi }}{7} + k2\pi \\x = \frac{{6\pi }}{7} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}\;c)\;sin4x - cos\left( {x + \frac{\pi }{6}} \right) = 0\\ \Leftrightarrow sin4x = cos\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{2} - x - \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{3} - x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{3} - x + k2\pi \\4x = \pi  - \frac{\pi }{3} + x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{15}} + k\frac{{2\pi }}{5}\\x = \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)

gợi ý: a)chia 2 vế cho căn 5

đặt \(\frac{1}{\sqrt{5}}=cosa\Rightarrow\frac{2}{\sqrt{5}}=sina\)

khi đó pt <=>sin(x-a)=\(\frac{3}{\sqrt{5}}>1\)

->vô nghiệm

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

b) \(\cos 3x - \sin 5x = 0\;\;\;\; \Leftrightarrow \cos 3x = \sin 5x\;\;\;\; \Leftrightarrow \cos 3x = \cos \left( {\frac{\pi }{2} - 5x} \right)\;\;\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \frac{\pi }{2} - 5x + k2\pi }\\{3x =  - \frac{\pi }{2} + 5x + k2\pi }\end{array}} \right.\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{8x = \frac{\pi }{2} + k2\pi }\\{ - 2x =  - \frac{\pi }{2} + k2\pi }\end{array}} \right.\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}}\\{x = \frac{\pi }{4} - k\pi }\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

sin x + cos x = 1 + sin x.cos x

⇔ sin x.cos x – sin x – cos x + 1 = 0

⇔ (sinx. cosx –sinx)- (cosx -1 ) =0

⇔ sinx. (cosx – 1) – (cosx -1) = 0

⇔ (sin x – 1)(cos x – 1) = 0

Vậy phương trình có tập nghiệm 

\(\Leftrightarrow\frac{\cos^2x-4\sin^2x.\cos^2x}{4\cos^2x}=\frac{1}{2}\left(\cos\frac{\pi}{3}-\cos2x\right)\)

\(\Leftrightarrow1-4\sin^2x=2\left(\frac{1}{2}-\cos2x\right)\)

\(\Leftrightarrow1-4\sin^2x=1-2\cos2x\)

\(\Leftrightarrow2\sin^2x=\cos2x\)

\(\Leftrightarrow1-\cos2x=\cos2x\)

\(\Leftrightarrow\cos2x=\frac{1}{2}\Leftrightarrow x=\pm\frac{\pi}{6}+k\pi,k\in Z\) thỏa mãn điều kiện