Question

determine whether the statement is true or false. Justify your answer. 99. The equation 2 sin 4t − 1 = 0 has four times the number of solutions in the interval [0, 2pi) as the equation 2 sin t − 1 = 0.

Equations and inequalities
ANSWERED
asked 2021-01-10
determine whether the statement is true or false. Justify your answer.
99. The equation \(\displaystyle{2}{\sin{{4}}}{t}−{1}={0}\) has four times the number of solutions in the interval \(\displaystyle{\left[{0},{2}\pi\right)}\) as the equation \(\displaystyle{2}{\sin{{t}}}−{1}={0}\).

Answers (1)

2021-01-11

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)
\(\displaystyle{2}{\sin{{t}}}-{1}={0}\)
\(\displaystyle{\sin{{t}}}={\frac{{{2}}}{{{2}}}}\)
\(\displaystyle{t}={{\sin}^{{-{1}}}{\left({\frac{{{1}}}{{{2}}}}\right)}}\)
\(\displaystyle{t}={\frac{{\pi}}{{{6}}}}\)

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