Ora,
\n$$\\begin{array}{*{20}{l}}
\n{\\operatorname{sen} \\left( {\\frac{\\pi }{6}\\left( {t + T} \\right) – \\frac{{5\\pi }}{3}} \\right) = \\operatorname{sen} \\left( {\\frac{\\pi }{6}t – \\frac{{5\\pi }}{3}} \\right)}& \\Leftrightarrow &{\\operatorname{sen} \\left( {\\frac{\\pi }{6}t + \\frac{\\pi }{6}T – \\frac{{5\\pi }}{3}} \\right) = \\operatorname{sen} \\left( {\\frac{\\pi }{6}t – \\frac{{5\\pi }}{3}} \\right)} \\\\
\n{}& \\Leftrightarrow &{\\frac{\\pi }{6}T = 2k\\pi ,k \\in \\mathbb{Z}} \\\\
\n{}& \\Leftrightarrow &{T = 12k,k \\in \\mathbb{Z}}
\n\\end{array}$$<\/p>\nPortanto, o per\u00edodo positivo m\u00ednimo da fun\u00e7\u00e3o $M$ \u00e9 $T = 12$ horas.<\/p>\n
Da\u00ed, no intervalo $\\left[ {0,24} \\right]$, a fun\u00e7\u00e3o $M$ alcan\u00e7ou os seguintes valores m\u00ednimo e m\u00e1ximo, respetivamente: ${M_{m\u00edn}} = 4,5 \\times ( – 1) + 7,5 = 3$ e ${M_{m\u00e1x}} = 4,5 \\times 1 + 7,5 = 12$, em metros.<\/p>\n
Portanto, nesse dia, $3 \\leqslant M \\leqslant 12$, em metros.
\n\u00ad<\/li>\n
Como $$\\begin{array}{*{20}{l}}
\n{M(t) = 3 \\wedge t \\in \\left[ {0,24} \\right]}& \\Leftrightarrow &{\\operatorname{sen} \\left( {\\frac{\\pi }{6}t – \\frac{{5\\pi }}{3}} \\right) =\u00a0 – 1 \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{\\frac{\\pi }{6}t – \\frac{{5\\pi }}{3} = \\frac{{3\\pi }}{2} + 2k\\pi ,k \\in \\mathbb{Z} \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{\\frac{t}{6} – \\frac{5}{3} = \\frac{3}{2} + 2k,k \\in \\mathbb{Z} \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{t = 19 + 12k,k \\in \\mathbb{Z} \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{t = 7}& \\vee &{t = 19}
\n\\end{array}}
\n\\end{array}$$
\na baixa-mar ocorreu \u00e0s 7 \u00e9 \u00e0s 19 horas desse dia.<\/p>\nComo $$\\begin{array}{*{20}{l}}
\n{M(t) = 12 \\wedge t \\in \\left[ {0,24} \\right]}& \\Leftrightarrow &{\\operatorname{sen} \\left( {\\frac{\\pi }{6}t – \\frac{{5\\pi }}{3}} \\right) = 1 \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{\\frac{\\pi }{6}t – \\frac{{5\\pi }}{3} = \\frac{\\pi }{2} + 2k\\pi ,k \\in \\mathbb{Z} \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{\\frac{t}{6} – \\frac{5}{3} = \\frac{1}{2} + 2k,k \\in \\mathbb{Z} \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{t = 13 + 12k,k \\in \\mathbb{Z} \\wedge t \\in \\left[ {0,24} \\right]} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{t = 1}& \\vee &{t = 13}
\n\\end{array}}
\n\\end{array}$$
\na preia-mar ocorreu \u00e0 1 e \u00e0s 13 horas desse dia.
\n\u00ad<\/li>\n