Problem 3
e sin x − e − sin x − 4 = 0
What is the number of real solution(s) of the equation above?
The answer is 0.
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6 solutions
Nice solution Paola ,Upvoted
e^sinx can have a maximum value of e^1=2.7 but in problem 4 and e^-sinx have been subtracted. which will give a negative value always since e^-sinx will be always positive. hence whole expression will be always negative and never be equal to zero..........
Let e s i n x = t . The equation becomes t 2 − 4 t − 1 = 0 .
⟹ t = 2 ± 5
So e s i n x = 2 + 5 o r 2 − 5
∵ e s i n x can never be less than 0 and its maximum value is e 1 < 2 + 5
Hence there is no real solution of the given equation.
Nice solution sir, upvoted
The graphical representation shows no real solution.
Lets see.
substitute
e
s
i
n
x
as
t
then the equation will be as foll:
t
2
−
4
t
−
1
=
0
Discriminant
=
2
5
therefore
t
=
(
4
±
2
5
)
÷
2
=
2
±
5
on comparing both values of t with e^sinx we find e^sinx can be -ve therefore e^sinx = 4+2\sqrt { 5 } now take both side ln -: \ln { { e }^{ sinx } } =\ln { \left( 2+\sqrt { 5 } \right) } \ sinx=\quad ln\left( 2+\sqrt { 5 } \right) on an approx value of 2+\sqrt{5} =4.1 then value of ln will be greater than 1 as approx value of e=2.7 nd sinx can never be more than one hence no solutions!!! therefore answer is 0
Moderator note:
Good solution.
A few minor typos like "we find e^sinx cannot be -ve"
FYI To type in Latex, you just need to use the brackets \ ( \ ) . I've edited the start of your solution, so you can refer to it.
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the thing you edited it should be e^sinx and not e^x = t if not then why??
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I only added brackets around your code. As I said, there were some minor typos in your solution. You can edit your solution by clicking on the edit button.
the thing you edited it should be e^sinx and not e^x = t if not then why??
we find e^sinx cannot be -ve is true.
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Right, what you have is "we find e^sinx can be -ve", in the first line of the second paragraph. It should be "cannot".
Easier than that. Rewrite the equation as: sinh(sin(x)) = 2
As sinh is an increasing function and sin(x) is between -1 and 1, sinh(sin(x)) is between sinh(-1) and sinh(1). sinh(-1) = -1.175 .... and sinh(1) = 1.175., therefore the equation is impossible.
Or, you could be a smart ass.
Just look into the question, no calculations. As the sin(x) is periodic, either the solution is zero, or infinite repetitions, and there is no way to represent infinite (not given) in this question.
:)
e sin x − e − sin x − 4 = e sin x − e sin x 1 − 4 = ( e sin x ) 2 − 4 ( e sin x ) − 1 = 0
Let m = e sin x ⇒ m 2 − 4 m − 1 = 0
Solve the quadratic ecuation we get it
m = 2 + 5 , 2 − 5 ⇒ sin x = ln 2 + 5 and sin x = ln 2 − 5
But ln 2 − 5 does not exist and ln 2 + 5 is bigger than one ∴ the equation has no real solutions.