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Mathematics
Mathematics
Problem
Solve
sin
(
θ
−
π
6
)
=
1
2
\sin\left(\theta - \dfrac{\pi}{6}\right) = \dfrac{1}{2}
sin
(
θ
−
6
π
)
=
2
1
for
θ
\theta
θ
in the interval
0
≤
θ
<
2
π
0\le \theta < 2\pi
0
≤
θ
<
2
π
. a)
{
π
6
,
5
π
6
}
\left\{\dfrac{\pi}{6},\;\dfrac{5\pi}{6}\right\}
{
6
π
,
6
5
π
}
b)
{
π
3
,
π
}
\left\{\dfrac{\pi}{3},\;\pi\right\}
{
3
π
,
π
}
c)
{
0
,
2
π
3
}
\left\{0,\;\dfrac{2\pi}{3}\right\}
{
0
,
3
2
π
}
d)
{
π
3
,
4
π
3
}
\left\{\dfrac{\pi}{3},\;\dfrac{4\pi}{3}\right\}
{
3
π
,
3
4
π
}
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