When you you found the vertex, I was hoping you would get the interval [-1, 0] which is the only place the parabola is lesser/equal than 1. Meanwhile, the sin is bigger than 3/4 only on a subinterval of (pi/3, 2pi/3). Clearly, since both intervals doesn't intersect each other, you find no solution.
we see that x^2+x+1>=|x|>=|sinx|>=sinx for every real number x.now the first equality holds for x=-1 and the second for x=0.hence x^2+x+1>sinx so there is no solution
Trying to graph in my head, and knowing sin[x] peaks at 1, then the +1 makes me think there will be no intersection at least for positive x. Now watching to see ….
This doesn't help. The question asks "where do these two lines meet?". If you differentiate each side and equate it you are looking at the slope of those graph at every value of x and you'd be asking "for which value of x do these graph have the same slope?", but two functions that cross do not have to have the same slope at that point (indeed often they don't) and vice verse two lines could never meet but they could have the same slope at a given point. Differentiating both sides and equating them again would be like asking "at which value of x does the rate of change of the slope of each function match?". The fact there are no real solutions just means that sin(x) is never increasing/decreasing in steepness as much as x^2+x+1
To add to the comment above, I would like to point out the OPs method makes sense for identities. E.g. differentiating sin^2 x + cos^2 x = 1 yields another identity. Reason being the two functions on the lhs and rhs coincide pointwise, therefore the slopes and thereafters coincide
What you have done is demonstrated that there is no inflection point. It does not help show whether there is a solution to the original problem. A counter example would be where the right hand side was x^2+x-1, which does have two solutions, but the double derivative is the same.
We have log and product log. I will define the sine log as S(sin(x) - x^2 - x) = 1. Therefore the solution is simply x = S(1). Q E.D.
Or, simpler, we have e and pi, so just define - say - q as a solution to the equation. Then x = q is a solution.
Yes, but how do you find complex solutions?
Substituting sinx = Im(e^ix) can help then just taking log will suffice
@@bixxuxboxplayz4986do you mean “ln(e^ix)”?
Sin(x) = (e^(ix) - e^(-ix))/2i
(e^(ix) - e^(-ix))/2i = x^2 + x + 1
Now we can solve it easily.
When you you found the vertex, I was hoping you would get the interval [-1, 0] which is the only place the parabola is lesser/equal than 1.
Meanwhile, the sin is bigger than 3/4 only on a subinterval of (pi/3, 2pi/3).
Clearly, since both intervals doesn't intersect each other, you find no solution.
we see that x^2+x+1>=|x|>=|sinx|>=sinx for every real number x.now the first equality holds for x=-1 and the second for x=0.hence x^2+x+1>sinx so there is no solution
Why do these inequalities work?
Trying to graph in my head, and knowing sin[x] peaks at 1, then the +1 makes me think there will be no intersection at least for positive x. Now watching to see ….
Interesting!
If it were cosx instead of sinx we would obtain solutions x=0 and x=-0.675 approximately. I used Newton's method to numerically find the solutions.
Looking at the graphs of functions cosx and x^2+x+1 you clearly see there are two intersecting points
👍
sin (X)=x^2 , even this seemly easy question . the 2nd answer is algebraic impossible.
are these equations called transcendental equations?
yes
So they can be solved by Regula-Falsi or Newton-Raphson methods
We can make use of the |sinx|
if you differentiate twice on both sides you get sinx = -2 which means no real solutions to x
This doesn't help. The question asks "where do these two lines meet?". If you differentiate each side and equate it you are looking at the slope of those graph at every value of x and you'd be asking "for which value of x do these graph have the same slope?", but two functions that cross do not have to have the same slope at that point (indeed often they don't) and vice verse two lines could never meet but they could have the same slope at a given point. Differentiating both sides and equating them again would be like asking "at which value of x does the rate of change of the slope of each function match?". The fact there are no real solutions just means that sin(x) is never increasing/decreasing in steepness as much as x^2+x+1
Good point! Thank you
To add to the comment above, I would like to point out the OPs method makes sense for identities. E.g. differentiating sin^2 x + cos^2 x = 1 yields another identity. Reason being the two functions on the lhs and rhs coincide pointwise, therefore the slopes and thereafters coincide
What you have done is demonstrated that there is no inflection point. It does not help show whether there is a solution to the original problem. A counter example would be where the right hand side was x^2+x-1, which does have two solutions, but the double derivative is the same.
Very illegal solution !
Suppose .. cosx = x² + x + 1
Differentiate twice
- cosx = 2
But... X = 0 and x= something satisfy.