1. Simplify the function

Given $f(x)=\left(1-\cos^2 x\right)(\lambda+\sin x), \qquad x\in\left(-\frac\pi2,\frac\pi2\right).$ Since $1-\cos^2 x=\sin^2 x,$ we get $f(x)=\sin^2 x(\lambda+\sin x).$

Let $t=\sin x.$ Because $x\in\left(-\frac\pi2,\frac\pi2\right)$, we have $t\in(-1,1)$ and $t=\sin x$ is strictly increasing. Hence maxima/minima of $f$ as a function of $x$ correspond exactly to maxima/minima of $g(t)=t^2(\lambda+t)=\lambda t^2+t^3, \qquad t\in(-1,1).$

So we study $g(t)$ on $(-1,1)$.

2. Find critical points

Differentiate: $g'(t)=2\lambda t+3t^2=t(2\lambda+3t).$ Thus critical points are $t=0, \qquad t=-\frac{2\lambda}{3}.$

For the function to have exactly one maximum and exactly one minimum, we need two distinct critical points inside $(-1,1)$.

• $t=0$ is always in $(-1,1)$.
• The second critical point must satisfy $-1< -\frac{2\lambda}{3}<1 \iff |\lambda|<\frac32.$
• Also it must be distinct from $0$, so $\lambda\ne 0.$

Hence necessary condition: $|\lambda|<\frac32, \quad \lambda\ne 0.$

3. Check nature of critical points

Second derivative: $g''(t)=2\lambda+6t.$

At $t=0$: $g''(0)=2\lambda.$

• If $\lambda>0$, then $g''(0)>0$, so $t=0$ is a local minimum.
• If $\lambda<0$, then $g''(0)<0$, so $t=0$ is a local maximum.

At $t=-\frac{2\lambda}{3}$: $g''\left(-\frac{2\lambda}{3}\right)=2\lambda+6\left(-\frac{2\lambda}{3}\right)=2\lambda-4\lambda=-2\lambda.$

• If $\lambda>0$, this is negative, so this point is a local maximum.
• If $\lambda<0$, this is positive, so this point is a local minimum.

Thus whenever $|\lambda|<\frac32 \text{ and } \lambda\ne 0,$ there is exactly one local maximum and exactly one local minimum.

4. Exclude boundary/degenerate cases

• If $\lambda=0$, then $g'(t)=3t^2,$ so only one stationary point $t=0$, not one maximum and one minimum.
• If $|\lambda|=\frac32$, then second critical point is at $t=\pm1$, which is not in the open interval $(-1,1)$.
• If $|\lambda|>\frac32$, then only one critical point lies in $(-1,1)$.

So the required set is $\left(-\frac32,\frac32\right)\setminus\{0\}.$

5. Match with options

This is Option A.

$\boxed{\left(-\frac32,\frac32\right)-\{0\}}$