If $\sin (2 x)=\frac{\sqrt{5}-1}{4}$ then $x=\frac{n}{2} \pi+(-1)^n(m)$, $n \in \mathbf{Z}$, find $m$.
If $\sin (2 x)=\frac{\sqrt{5}-1}{4}$ then $x=\frac{n}{2} \pi+(-1)^n(m)$, $n \in \mathbf{Z}$, find $m$.
$\frac{\pi}{10}$
$\frac{\pi}{5}$
$\frac{\pi}{20}$
$\frac{\pi}{40}$
Solution
$
\text { } \begin{aligned}
\sin 2 x & =\frac{\sqrt{5}-1}{4} \\
\sin 2 x & =\sin \frac{\pi}{10} \\
2 x & =n \pi+(-1)^n \frac{\pi}{10} \\
x & =\frac{n \pi}{2}+(-1)^n \frac{\pi}{20}
\end{aligned}
$
but given,
$
x=\frac{n \pi}{2}+(-1)^n \cdot m
$
On comparison, we get $m=\frac{\pi}{20}$ Hence, option (3) is correct