Can $f(x)=\sin(x)  +  \cos^2(x)$ take the value $\sqrt{2}$? Can $f(x)$ take the value of $\sqrt{2}$, where
 $$f(x)  = \sin(x)  +  (\cos(x))^2\quad ?$$
When equating the value of $f(x)$ to $\sqrt{2}$, it gives imaginary values of $\sin(x)$.
Thanks in advance.
 A: The local extrema will occur where $f\,'(x)=0$, or when factored, $(\cos x)(1-2\sin x)=0$. We can use the identity $\sin^2 x+\cos^2x=1$ to compute the possible extrema as
$$\cos =0 \implies \qquad \pm\sqrt{1-0^2}+0^2=\pm1 $$
$$\sin = 1/2 \implies \qquad \frac{1}{2}+\left|1-\frac{1}{2^2}\right|=5/4.$$
Of course $f$ is periodic and bounded so its image must be $(-1,5/4)$, which does not include $\sqrt{2}$.
$\hskip 1.6in$ 
This reasoning applies to the reals. Allowing complex arguments is another story.
A: You ask if $f(x) = \sin x + \cos ^2 x$ can be equal to $\sqrt 2$. In short, no, it can't.
Now is when I take a moment to thank anon for noting that $1 < \sqrt 2$. That whole arithmetic thing.... yeah. Anyhow.
If one uses the first derivative test, then you can note that the max should occur around $\pi/6$ or $5 \pi / 6$ (plus multiples of $2\pi$ if desired). At both of these values, the function is a mere $1.25$, which (I hope anon checks me again) is less than $\sqrt 2$.
A: Since $\cos^2x=1-\sin^2 x$, our equation becomes $\sin^2 x-\sin x+\sqrt{2}-1=0$.
Solve the equation $w^2-w+\sqrt{2}-1=0$. The roots are 
$$w=\frac{1\pm \sqrt{1-4(\sqrt{2}-1})}{2}.$$
The discriminant $1-4(\sqrt{2}-1)$ is negative, so our quadratic in $w$ has no real roots. It follows that there cannot be a real number $x$ such that $\sin x+\cos^2 x=\sqrt{2}$.
This does not fully answer the question. We can ask whether there are complex non-real solutions.  For that, look at the two non-real values of $w$ obtained above. Call them $w_1$ and $w_2$. Recall that if $z$ is a complex number, then
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}.$$
So we want to solve the equations
$$e^{iz}-e^{-iz}=2iw_j \quad (j=1,2).$$
Multiply both sides by $e^{iz}$, and simplify. We get
$$e^{2iz}-2iw_je^{iz}-1=0.$$
Make the substitution $u=e^{iz}$. We arrive at the equations
$$u^2-2iw_j u-1=0 \quad (j=1,2).$$
These are a quadratic equation with complex coefficients. One can write down the solutions in more or less the usual way, and from them obtain the values of $z$ that work. There are  infinitely many of them, just like there are, for example, infinitely many $x$ such that $\sin x=1/2$.
Remark To show that there are no real solutions, one does not need to recall the Quadratic Formula. We are interested in the equation  $\sin^2 x-\sin x-1+\sqrt{2}=0$.
If we complete the square, we get
$$\sin^2 x-\sin x-1=\left(\sin x-\frac{1}{2}\right)^2-\frac{5}{4}.$$
Since $\left(\sin x-\frac{1}{2}\right)^2\ge 0$, it follows that 
$\sin^2 x-\sin x-1$ must always be $-\frac{5}{4}$. Since $\sqrt{2}-\frac{5}{4}$ is positive, we conclude that we cannot have $\sin^2 x-\sin x-1+\sqrt{2}=0$.
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A: Put $\sin x =:u$. Then the question is whether the function
$$g(u):=u+(1-u^2)={5\over4}-\Bigl(u-{1\over2}\Bigr)^2$$
can take the value $\sqrt{2}$ in the interval $-1\leq u\leq 1$. This is obviously not the case.
