$$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right)$$

$$ \Rightarrow $$ f(x) = sin² x($$\lambda $$ + sinx) ....(1)

$$ \therefore $$ f'(x) = 2sinx cosx ($$\lambda $$ +sinx) + sin²x (cosx)

$$ \Rightarrow $$ f'(x) = sin2x($${{2\lambda + 3\sin x} \over 2}$$)

For maixma and minima, f'(x) = 0

$$ \therefore $$ sin2x = 0 or 2$$\lambda $$ + 3sinx = 0

when sin2x = 0 $$ \Rightarrow $$ x = 0

or when 2$$\lambda $$ + 3sinx = 0

$$ \Rightarrow $$ sin x = $$ - {{2\lambda } \over 3}$$

As $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$

$$ \therefore $$ -1 < sinx < 1

$$ \Rightarrow $$ -1 < $$ - {{2\lambda } \over 3}$$ < 1

$$ \Rightarrow $$ $$ - {3 \over 2}$$ < $$\lambda $$ < $$ {3 \over 2}$$

$$ \therefore $$ $$\lambda $$ $$ \in $$ $$\left( { - {3 \over 2},{3 \over 2}} \right)$$ - {0}

Note : If $$\lambda $$ = 0 $$ \Rightarrow $$ f(x) = sin³x [from (1)]

Which is monotonic. so no maxima/minima.