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
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 $$\displaystyle{2}{\sin{{t}}}−{1}={0}$$.

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