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Mathematics
Mathematics
Problem
The solutions of
sin
2
θ
=
3
2
\sin 2\theta = \dfrac{\sqrt{3}}{2}
sin
2
θ
=
2
3
in the interval
0
≤
θ
<
2
π
0 \leq \theta < 2\pi
0
≤
θ
<
2
π
are a)
{
π
3
,
2
π
3
,
4
π
3
,
5
π
3
}
\left\{\dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}\right\}
{
3
π
,
3
2
π
,
3
4
π
,
3
5
π
}
b)
{
π
4
,
3
π
4
,
5
π
4
,
7
π
4
}
\left\{\dfrac{\pi}{4},\; \dfrac{3\pi}{4},\; \dfrac{5\pi}{4},\; \dfrac{7\pi}{4}\right\}
{
4
π
,
4
3
π
,
4
5
π
,
4
7
π
}
c)
{
π
6
,
π
3
,
7
π
6
,
4
π
3
}
\left\{\dfrac{\pi}{6},\; \dfrac{\pi}{3},\; \dfrac{7\pi}{6},\; \dfrac{4\pi}{3}\right\}
{
6
π
,
3
π
,
6
7
π
,
3
4
π
}
d)
{
π
6
,
5
π
6
,
7
π
6
,
11
π
6
}
\left\{\dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}\right\}
{
6
π
,
6
5
π
,
6
7
π
,
6
11
π
}
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