find the value of theeta when 2cos theeta = 3sin theeta
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Explanation:
Substitute
1
−
sin
2
(
θ
)
for
cos
2
(
θ
)
:
2
(
1
−
sin
2
(
θ
)
)
+
3
sin
(
θ
)
=
0
Use the distributive property:
2
−
2
sin
2
(
θ
)
)
+
3
sin
(
θ
)
=
0
Multiply both side by -1:
2
sin
2
(
θ
)
)
−
3
sin
(
θ
)
−
2
=
0
This is a quadratic where the variable is
sin
(
θ
)
.
It looks like it will factor:
#(sin(theta) - 2)(2sin(theta) + 1) = 0
sin
(
θ
)
=
2
and
sin
(
θ
)
=
−
1
2
We must discard the first root, because it is outside the range of the sine function.
Turning our attention to the second root:
sin
(
θ
)
=
−
1
2
Rotating counterclockwise from 0, the first encounter of this is at:
θ
=
7
π
6
The next encounter with this is at:
θ
=
11
π
6
Add integer rotations of
2
π
to both:
θ
=
7
π
6
+
2
n
π
and
θ
=
11
π
6
+
2
n
π
Where n can be any integer (positive, negative, or zero)
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