How many real roots does the equation e^{sin | vedclass

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(C) Let $u = e^{\sin x}$. Since $-1 \le \sin x \le 1$,the range of $u$ is $[e^{-1}, e^1]$,which is approximately $[0.368, 2.718]$.Substituting $u$ int

Vedclass · Accepted Answer

No real roots.

Vedclass · Answer

(C) Let $u = e^{\sin x}$. Since $-1 \le \sin x \le 1$,the range of $u$ is $[e^{-1}, e^1]$,which is approximately $[0.368, 2.718]$.Substituting $u$ into the equation,we get $u - \frac{1}{u} - 4 = 0$.Multiplying by $u$,we obtain the quadratic equation $u^2 - 4u - 1 = 0$.Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $u = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$.Since $u = 2 + \sqrt{5} \approx 4.236$ and $u = 2 - \sqrt{5} \approx -0.236$,neither of these values falls within the range $[0.368, 2.718]$.Therefore,there are no real values of $x$ that satisfy the equation.Thus,the equation has no real roots.

How many real roots does the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$ have?

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