Given

\(\displaystyle{2}{\sin{{4}}}{t}-{1}={0}---{\left({1}\right)}\)

\(\displaystyle{2}{\sin{{t}}}-{1}={0}---{\left({2}\right)}\)

FALSE

EXPLAINATION

From eq(1)

\(\displaystyle{2}{\sin{{4}}}{t}-{1}={0}\)

\(\displaystyle{\sin{{4}}}{t}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{4}{t}={{\sin}^{{-{1}}}{\left({\frac{{{1}}}{{{2}}}}\right)}}\)

\(\displaystyle{t}={\frac{{{1}}}{{{4}}}}(\frac{\pi}{6})\)

From eq(2)

\(\displaystyle{2}{\sin{{t}}}-{1}={0}\)

\(\displaystyle{\sin{{t}}}={\frac{{{2}}}{{{2}}}}\)

\(\displaystyle{t}={{\sin}^{{-{1}}}{\left({\frac{{{1}}}{{{2}}}}\right)}}\)

\(\displaystyle{t}={\frac{{\pi}}{{{6}}}}\)