Question

Range of 2cos^2x+sin^2x

Answer 1

I found:
x
=
0
,
π
,
2
π

x
=
π
6
,
5
6
π

Explanation:
We can use the fact that:
cos
2
(
x
)
=
1
−
sin
2
(
x
)

and write:
2
(
1
−
sin
2
(
x
)
)
+
sin
(
x
)
−
2
=
0

2
−
2
sin
2
(
x
)
+
sin
(
x
)
−
2
=
0

solve as a normal Quadratic Equation but in
sin
(
x
)
instead of
x
:
collecting
sin
(
x
)
:
sin
(
x
)
[
−
2
sin
(
x
)
+
1
]
=
0

we get 2 solutions:
sin
(
x
)
=
0

and:
−
2
sin
(
x
)
+
1
=
0

so that must be:
sin
(
x
)
=
0
when
x
=
0
,
π
,
2
π

sin
(
x
)
=
1
2
when
x
=
π
6
,
5
6
π