Find the number of solutions of the equation tan x + sec x = 2 cos x; cos x ≠ 0, lying in the interval (0, π).
Answers
Answer:
This equation has No solution according to the given Constraints.
Explanation:
Given that
tan x + sec x = 2 cos x
Given constraints are
- cos x ≠ 0
- lying in the interval (0, π)
Now
tan x + sec x = 2 cos x
sin x/cos x + 1/cos x = 2 cos x
Multiplying both sides by cos x , we get
sin x + 1 = 2 cos² x
sin x + 1 = 2 ( 1 - sin² x )
sin x + 1 = 2 - 2 sin² x
sin x + 1 - 2 + 2 sin² x = 0
2 sin² x + sin x - 1 = 0
2 sin² x + 2 sin x - sin x - 1 = 0
2 sin x ( sin x + 1 ) - 1 (sin x + 1) = 0
( sin x + 1 ) ( 2 sin x - 1 ) = 0
⇒ sin x + 1 = 0 , 2 sin x - 1 = 0
For sin x + 1 = 0 :
sin x + 1 = 0
sin x = - 1
x = 3π/2 , which is not valid. Because it is not in (0, π)
For 2 sin x - 1 = 0 :
2 sin x - 1 = 0
2 sin x = 1
sin x = 1/2
x = π/2 , which is nit valid. Because cosπ/2 = 0 And given that cos x ≠ 0
So this equation has No solution according to the given Constraints.
Hope it will help YOU.