2.
sin-1 x + cos2x, xel-1, 1) =
Answers
Answer:
Step-by-step explanation:
Given
1+cos2x
=
2
sin
−1
(sinx)
1+2cos
2
x−1
=
2
sin
−1
(sinx)
⇒
2cos
2
x
=
2
sin
−1
(sinx)
1. For
2
−π
≤x≤
2
π
2cos
2
x
=
2
sin
−1
(sinx)
→
2
cosx=
2
x
→cosx=x
Graph of y=cosx and y=x intersects only 1 time for
2
−π
≤x≤
2
π
Thus we get one real solution from here .
2.For x∈[−π,
2
−π
)
⇒
2cos
2
x
=
2
sin
−1
(sinx)
⇒−cosx=x+π
Graph of y=-cosx and y=x+π do not intersect forx∈[−π,
2
−π
) .
Thus we get no real solution from here .
3. For x∈(
2
π
,π]
⇒
2cos
2
x
=
2
sin
−1
(sinx)
⇒−cosx=π−x
Graph of y=-cosx and y=π−x intersect only 1 time for x∈[−π,
2
−π
) .
Thus we get one real solution from here .