` Wiskundeleraar

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Opdracht 1

Los exact op:

1. $2\sin \left( {4t} \right) - 3 = - 4$
2. $\cos ^2 \left( {2t} \right) = \frac{1}{2}$
3. $\sin \left( {2t + \pi } \right) \cdot \cos \left( {3t- \pi } \right) = 0$
4. \eqalign{\sin \left( {\frac{1}{4}\pi x} \right) = \sin \left( {\frac{1}{4}\pi } \right)}

Uitwerkingen

a.
\eqalign{ & 2\sin \left( {4t} \right) - 3 = - 4 \cr & 2\sin \left( {4t} \right) = - 1 \cr & \sin \left( {4t} \right) = - \frac{1} {2} \cr & 4t = - \frac{1} {6}\pi + k \cdot 2\pi \vee 4t = 1\frac{1} {6}\pi + k \cdot 2\pi \cr & 4t = 1\frac{5} {6}\pi + k \cdot 2\pi \vee 4t = 1\frac{1} {6}\pi + k \cdot 2\pi \cr & t = \frac{{11}} {{24}}\pi + k \cdot \frac{1} {4}\pi \vee t = \frac{7} {{24}}\pi + k \cdot \frac{1} {2}\pi \cr}
b.
\eqalign{ & \cos ^2 \left( {2t} \right) = \frac{1} {2} \cr & \cos \left( {2t} \right) = - \sqrt {\frac{1} {2}} \vee \cos \left( {2t} \right) = \sqrt {\frac{1} {2}} \cr & \cos \left( {2t} \right) = - \frac{1} {2}\sqrt 2 \vee \cos \left( {2t} \right) = \frac{1} {2}\sqrt 2 \cr & 2t = \frac{3} {4}\pi + k \cdot 2\pi \vee 2t = 1\frac{1} {4}\pi + k \cdot 2\pi \vee 2t = \frac{1} {4}\pi + k \cdot 2\pi \vee 2t = 1\frac{3} {4}\pi + k \cdot 2\pi \cr & t = \frac{3} {8}\pi + k \cdot \pi \vee t = \frac{5} {8}\pi + k \cdot \pi \vee t = \frac{1} {8}\pi + k \cdot \pi \vee t = \frac{7} {8}\pi + k \cdot \pi \cr & t = \frac{1} {8}\pi + k \cdot \frac{1} {4}\pi \cr}
c.

\eqalign{ & \sin \left( {2t + \pi } \right) \cdot \cos \left( {3t - \pi } \right) = 0 \cr & \sin (2t + \pi ) = 0 \vee \cos (3t - \pi ) = 0 \cr & 2t + \pi = k \cdot \pi \vee 3t - \pi = \frac{1} {2}\pi + k \cdot \pi \cr & 2t = k \cdot \pi \vee 3t = 1\frac{1} {2}\pi + k \cdot \pi \cr & t = k \cdot \frac{1} {2}\pi \vee t = \frac{1} {2}\pi + k \cdot \frac{1} {3}\pi \cr}

Naschrift
Bij de oplossing hierboven zitten nog een aantal dubbelen...

d.
\eqalign{ & \sin \left( {\frac{1} {4}\pi t} \right) = \sin \left( {\frac{1} {4}\pi } \right) \cr & \frac{1} {4}\pi t = \frac{1} {4}\pi + k \cdot 2\pi \vee \frac{1} {4}\pi t = \frac{3} {4}\pi + k \cdot 2\pi \cr & \pi t = \pi + k \cdot 8\pi \vee \pi t = 3\pi + k \cdot 8\pi \cr & t = 1 + k \cdot 8 \vee t = 3 + k \cdot 8 \cr}