f: \(cos7x-\sqrt3\cdot\sin7x-\sin x=\sqrt3\cdot cosx\)

=>\(\frac12\cdot cos7x-\frac{\sqrt3}{2}\cdot\sin7x=\frac12\cdot\sin x+\frac{\sqrt3}{2}\cdot cosx\)

=>\(\sin\left(\frac{\pi}{6}-7x\right)=\sin\left(x+\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}-7x+\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\ -7x+\frac{\pi}{6}=\pi-x-\frac{\pi}{3}+k2\pi=-x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}-7x-x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ -7x+x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}-8x=\frac{\pi}{6}+k2\pi\\ -6x=\frac12\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{48}-\frac{k\pi}{4}\\ x=-\frac{1}{12}\pi-\frac{k\pi}{3}\end{array}\right.\)

e: \(5\cdot\sin2x-6\cdot cos^2x=13\)

=>\(5\cdot\sin2x-6\cdot\frac{1+cos2x}{2}=13\)

=>\(5\cdot\sin2x-3-3\cdot cos2x=13\)

=>\(5\cdot\sin2x-3\cdot cos2x=16\)

Vì \(5^2+\left(-3\right)^2=25+9=34<16^2\)

nên phương trình vô nghiệm

a) Ta có : -\(\sqrt{a^2+b^2}< =asinx+bcosx< =\sqrt{a^2+b^2}\)

=> \(-\sqrt{12^2+\left(-5\right)^2}< =y< =\sqrt{12^2+\left(-5\right)^2}\)

<=> \(-\sqrt{13}< =y< =\sqrt{13}\)

Vậy min=\(-\sqrt{13}\) ,max=\(\sqrt{13}\)

b) \(-\sqrt{9+16}< =3cosx-4sinx< =\sqrt{9+16}\)

<=> -5 <=3cos x -4sinx <= 5

<=> 0<= y <= 10

Vậy min=0 max=10

tanx=sinx/cosx=5/3

=sin=3/5cos

=>2cosx-3(3/5cos)/3cosx+4(5/3cos)

=1/30

\(tanx=\dfrac{sinx}{cosx}\)

\(\Rightarrow M=\dfrac{2sinx}{\dfrac{cosx}{\dfrac{4sinx}{cosx}}}-\dfrac{3cosx}{\dfrac{cosx}{\dfrac{7cosx}{cosx}}}\)

\(M=\dfrac{2tanx-3}{4tanx+7}\)

\(M=\dfrac{2.\left(-2\right)-3}{4.2+7}\)

\(M=\dfrac{1}{15}\)