Question

Answer 1

\(sin18^0=cos72^0=2cos^236^0-1=2\left(1-2.sin^218^0\right)^2-1\)\(=1-8.sin^218^0+8.sin^418^0\)

Có \(8.sin^318^0+m.sin^218^0=1\)

\(\Rightarrow\left(sin18^0-1\right)\left(8.sin^318^0+m.sin^218^0-1\right)=0\) ( vì sin18⁰-1\(\ne\)0)

\(\Leftrightarrow8.sin^418^0+sin^318^0\left(m-8\right)-m.sin^218^0-sin18^0+1=0\)

\(\Leftrightarrow8.sin^418^0+sin^318^0\left(m-8\right)-m.sin^218^0-\left(1-8.sin^218^0+8.sin^418^0\right)+1=0\)

\(\Leftrightarrow sin^318^0\left(m-8\right)-sin^218^0\left(m-8\right)=0\)

\(\Leftrightarrow\left(m-8\right)\left(sin^318^0-sin^218^0\right)=0\)

\(\Rightarrow m=8\)