What is the maximum value of $4\sin^2(x) + 3\cos^2(x) + \sin\left(\dfrac{x}{2}\right) + \cos\left(\frac{x}{2}\right)$?

Question

What is the maximum value of the following function? $$4\sin^2(x) + 3\cos^2(x) + \sin\left(\dfrac{x}{2}\right) + \cos\left(\frac{x}{2}\right)$$

Here is some of my work.

I have tried to solve the problem in this way, but got an answer of $4$. The answer provided is $4+\sqrt{2}$.

I cannot find where I have made a mistake. Please find where I have made mistake in the image provided....

Sorry I have written x=pi/2 by mistake, but I have done it using x=0

Answer 1

$4\text{sin}^2x + 3\text{cos}^2x+ \text{sin}(\frac{x}{2}) + \text{cos}(\frac{x}{2})$

= $4\text{sin}^2x + 3-3\text{sin}^2x+ \text{sin}(\frac{x}{2}) + \text{cos}(\frac{x}{2})$

= $\text{sin}^2x + 3+ \text{sin}(\frac{x}{2}) + \text{cos}(\frac{x}{2})$

=$\text{sin}^2x + 3+\sqrt2 \text{sin}(\frac{\pi}{4}+\frac{x}{2})$

Now, when $x=\frac{\pi}{2}$, $\text{sin}(\frac{\pi}{4}+\frac{x}{2})=1$

So, when $x=\frac{\pi}{2}$, $\text{sin}^2x + 3+\sqrt2 \text{sin}(\frac{\pi}{4}+\frac{x}{2})=4+\sqrt2$

Answer 2

For example:

$$\sin\frac{x}{2}\neq\sqrt{1-\cos^2\frac{x}{2}}.$$ Try $x=-60^{\circ}.$

My solution.

By C-S $$4\sin^2x+3\cos^2x+\sin\frac{x}{2}+\cos\frac{x}{2}=4-\cos^2x+\sin\frac{x}{2}+\cos\frac{x}{2}\leq$$ $$\leq 4+\sqrt{2(\sin^2\frac{x}{2}+\cos^2\frac{x}{2})}=4+\sqrt2.$$ The equality occurs for $x=90^{\circ},$ which says that we got a maximal value.

The equality $$\sin\frac{x}{2}+\cos\frac{x}{2}\leq\sqrt{2(\sin^2\frac{x}{2}+\cos^2\frac{x}{2})}$$ we can prove also by the following way.

Let $\sin\frac{x}{2}=a$ and $\cos\frac{x}{2}=b$.

Thus, we need to prove that $$a+b\leq\sqrt{2(a^2+b^2)}.$$ Indeed, for $a+b\leq0$ it's obvious.

But for $a+b>0$ we need to prove that $$(a+b)^2\leq2(a^2+b^2)$$ or $$a^2+2ab+b^2\leq2a^2+2b^2$$ or $$(a-b)^2\geq0,$$ which is obvious again.