Question

[Lyly]

Giuap mk với  giải bt vs ạ 

Answer 1

[Capheny Bản Quyền]

cos2x.tan6x=sin10x 

ĐK : cos6x khác 0 

cos2x.sin6x/cos6x=sin10x

sin6xcos2x=sin10xcos6x 

1/2(sin8x+sin4x)=1/2(sin16x+sin4x)

sin8x+sin4x=sin16x+sin4x

sin8x=sin16x

sin16x=sin8x

\(\orbr{\begin{cases}16x=8x+k2pi\\16x=\frac{pi}{2}-8x+k2pi\end{cases}}\)   

\(\orbr{\begin{cases}16x-8x=k2pi\\16x+8x=\frac{pi}{2}+k2pi\end{cases}}\)   

\(\orbr{\begin{cases}8x=k2pi\\24x=\frac{pi}{2}+k2pi\end{cases}}\)   

\(\orbr{\begin{cases}x=\frac{kpi}{4}\\x=\frac{pi}{48}+\frac{kpi}{12}\end{cases}\left(k\in Z\right)}\)

Answer 2

[Capheny Bản Quyền]

sin^2(4x)+cos^2(6x)=1

\(\frac{1-cos8x}{2}+\frac{1+cos12x}{2}=1\)   

\(\frac{1}{2}-\frac{1}{2}cos8x+\frac{1}{2}+\frac{1}{2}cos12x=1\)   

\(\frac{1}{2}cos12x-\frac{1}{2}cos8x=0\)   

\(cos12x-cos8x=0\)   

\(-2sin10xsin2x=0\)   

\(\orbr{\begin{cases}sin10x=0\\sin2x=0\end{cases}}\)   

\(\orbr{\begin{cases}10x=kpi\\2x=kpi\end{cases}}\)   

\(\orbr{\begin{cases}x=\frac{kpi}{10}\\x=\frac{kpi}{2}\end{cases}}\)   

\(x=\frac{kpi}{10}\left(k\in Z\right)\)