Answer to Question #95332 in Real Analysis for Rajni

Question

Prove that between any two real roots of e^(3x) sin 5x =10?, there is at least one real root of e^(3x) cos5x+6=0 ?

Answer 1

Consider two functions $f(x)=e^{3x}sin{5x}-10$ and $g(x)=e^{3x}cos{5x}+6$

As value of $e^{3x}sin{5x}-10$ changes from positive to negative as we move from $x=1.5$ to $x=2$ and from $x=3$ to $x=3.5$

So there must be a real roots of $e^{3x}sin{5x}-10$ in the interval $(1.5, 2)$ and $(3, 3.5)$

Also the value of $e^{3x}cos{5x}+6$ changes from negative to positive as we move from $x=2.5$ to $x=3$

So there must be a real root of $e^{3x}cos{5x}+6$ in the interval $(2.5, 3)$

Hence between two real roots of $e^{3x}sin{5x}-10$ there exists at least one real root of $e^{3x}cos{5x}+6$