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Los exact op:

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

Uitwerkingen

$
\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}
$

$
\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}
$

$
\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...

$
\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}
$

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