2cos^2 theta
- 3cos theta +1=0
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0
Answer:
Explanation:
Let's factorise the LHS
2
cos
2
θ
−
3
cos
θ
+
1
=
(
2
cos
θ
−
1
)
(
cos
θ
−
1
)
=
0
Therefore,
2
cos
θ
−
1
=
0
cos
θ
=
1
2
and
cos
θ
−
1
=
0
cos
θ
=
1
We are looking for
θ
∈
[
0
,
2
π
[
When
cos
θ
=
1
2
θ
=
π
3
,
5
π
3
When
cos
θ
=
1
θ
=
0
Step-by-step explanation:
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