The set of all real values of $$\lambda$$ for which the function $$f(x) = (1 - \cos^2 x) \cdot (\lambda + \sin x)$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, has exactly one maxima and exactly one minima, is:

We have the function

$$f(x)=\bigl(1-\cos^2x\bigr)\,(\lambda+\sin x),\qquad x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right),\qquad\lambda\in\mathbb R.$$

Because the Pythagorean identity says $$\sin^2x+\cos^2x=1,$$ we may write

$$1-\cos^2x=\sin^2x.$$

Substituting this into the definition of $$f(x)$$ gives

$$f(x)=\sin^2x\;(\lambda+\sin x).$$

To locate local maxima and minima we set the first derivative equal to zero. First we differentiate. Write

$$g(x)=\sin^2x,\qquad h(x)=\lambda+\sin x,$$

so that $$f(x)=g(x)\,h(x).$$ Using the product rule $$\bigl(g\,h\bigr)'=g'h+gh',$$ and the facts

$$g'(x)=2\sin x\cos x=\sin 2x,\qquad h'(x)=\cos x,$$

we obtain

$$f'(x)=\sin 2x\;(\lambda+\sin x)+\sin^2x\;\cos x.$$

Replacing $$\sin 2x$$ by $$2\sin x\cos x$$ simplifies the expression:

$$f'(x)=2\sin x\cos x\;(\lambda+\sin x)+\sin^2x\cos x.$$

Both terms contain the factor $$\sin x\cos x,$$ so we factor it out:

$$f'(x)=\sin x\cos x\;\bigl[\,2(\lambda+\sin x)+\sin x\,\bigr].$$

Simplifying the bracket further,

$$2(\lambda+\sin x)+\sin x=2\lambda+2\sin x+\sin x=2\lambda+3\sin x.$$

Hence

$$f'(x)=\sin x\cos x\;\bigl(2\lambda+3\sin x\bigr).$$

Critical points arise when $$f'(x)=0,$$ i.e.

$$\sin x=0\quad\text{or}\quad\cos x=0\quad\text{or}\quad2\lambda+3\sin x=0.$$

Inside the open interval $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$ we note

• $$\cos x=0$$ only at $$x=\pm\dfrac{\pi}{2},$$ which are not in the domain, so they are discarded.

• $$\sin x=0$$ gives the critical point $$x_0=0.$$

• $$2\lambda+3\sin x=0$$ yields

$$\sin x=-\dfrac{2\lambda}{3}.$$

This equation has a real solution in the given interval precisely when

$$-1<-\dfrac{2\lambda}{3}<1.$$

Multiplying throughout by $$-\dfrac{3}{2}$$ (remembering to reverse the inequalities) gives

$$-\,\dfrac{3}{2}<\lambda<\dfrac{3}{2}.$$

If $$\lambda=0,$$ the value $$-\dfrac{2\lambda}{3}=0$$ coincides with the already found root $$x_0=0,$$ so we would have only one critical point. To obtain exactly two distinct critical points we therefore require

$$\lambda\in\left(-\dfrac{3}{2},\dfrac{3}{2}\right)\setminus\{0\}.$$

Let $$x_1$$ denote the additional root satisfying $$\sin x_1=-\dfrac{2\lambda}{3}\neq0.$$ The two critical points are thus

$$x_1\quad\text{and}\quad x_0=0.$$

We now examine the nature (maxima or minima) of these points by studying the sign of $$f'(x).$$ The factorisation

$$f'(x)=\bigl(\sin x\bigr)\bigl(\cos x\bigr)\bigl(2\lambda+3\sin x\bigr)$$

makes the sign analysis straightforward because $$\cos x>0$$ throughout the interval. Only the signs of $$\sin x$$ and $$2\lambda+3\sin x$$ matter.

Case 1: $$\lambda>0.$$

Then $$-\dfrac{2\lambda}{3}<0,$$ so $$x_1<0.$$ Moving from left to right along the interval we have

• $$( -\dfrac{\pi}{2},\,x_1):\quad\sin x<0,\;2\lambda+3\sin x<0\;\Longrightarrow\;f'(x)>0;$$
• $$x=x_1:\;f'(x)=0;$$
• $$(x_1,\,0):\quad\sin x<0,\;2\lambda+3\sin x>0\;\Longrightarrow\;f'(x)<0;$$
• $$x=0:\;f'(x)=0;$$
• $$(0,\,\dfrac{\pi}{2}):\quad\sin x>0,\;2\lambda+3\sin x>0\;\Longrightarrow\;f'(x)>0.$$

The derivative changes from positive to negative at $$x_1,$$ giving a local maximum there, and from negative to positive at $$x_0=0,$$ giving a local minimum. Thus we obtain exactly one maxima and one minima.

Case 2: $$\lambda<0.$$

Now $$-\dfrac{2\lambda}{3}>0,$$ so $$x_1>0.$$ The sign table becomes

• $$( -\dfrac{\pi}{2},\,0):\quad\sin x<0,\;2\lambda+3\sin x<0\;\Longrightarrow\;f'(x)>0;$$
• $$x=0:\;f'(x)=0;$$
• $$(0,\,x_1):\quad\sin x>0,\;2\lambda+3\sin x<0\;\Longrightarrow\;f'(x)<0;$$
• $$x=x_1:\;f'(x)=0;$$
• $$(x_1,\,\dfrac{\pi}{2}):\quad\sin x>0,\;2\lambda+3\sin x>0\;\Longrightarrow\;f'(x)>0.$$

Here the derivative switches from positive to negative at $$x_0=0,$$ producing a local maximum, and from negative to positive at $$x_1,$$ producing a local minimum. Again there is exactly one maxima and one minima.

Consequently, every $$\lambda$$ satisfying

$$\boxed{\,-\dfrac{3}{2}<\lambda<\dfrac{3}{2},\;\lambda\neq0\,}$$

gives exactly two distinct critical points, one a maximum and the other a minimum. No other value of $$\lambda$$ works:

• If $$\lambda=\pm\dfrac{3}{2},$$ the second root would correspond to $$\sin x=\mp1,$$ i.e. to $$x=\pm\dfrac{\pi}{2},$$ outside the open interval, leaving only a single critical point.

• If $$\lambda=0,$$ both conditions $$\sin x=0$$ and $$2\lambda+3\sin x=0$$ coincide, again giving only one critical point.

The required set is therefore

$$\left(-\dfrac{3}{2},\dfrac{3}{2}\right)\setminus\{0\}.$$

Hence, the correct answer is Option D.