The equation e^{sin x} - e^{sin(-x)} - 4 = 0 | vedclass

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(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$.Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,we have $t \in [e^{-1}, e^1]$,i.e.,$t \in [1

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no real roots

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(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$.Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,we have $t \in [e^{-1}, e^1]$,i.e.,$t \in [1/e, e]$.The equation becomes $t - \frac{1}{t} - 4 = 0$.Multiplying by $t$,we get $t^2 - 4t - 1 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5}$.Case $1$: $t = 2 - \sqrt{5}$. Since $\sqrt{5} \approx 2.236$,$2 - \sqrt{5} Case $2$: $t = 2 + \sqrt{5}$. Since $\sqrt{5} \approx 2.236$,$t \approx 4.236$. Since $e \approx 2.718$,$t > e$. However,the maximum value of $e^{\sin x}$ is $e^1 = e \approx 2.718$. Since $4.236 > 2.718$,there is no real value of $x$ such that $e^{\sin x} = 2 + \sqrt{5}$.Therefore,the equation has no real roots.

The equation $e^{\sin x} - e^{\sin(-x)} - 4 = 0$ has

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