maximum value of $\cos^4(x) + \sin^2 (x) + \cos (x)$, when $x \geq 0$ How do i solve this?
What is the maximum value of $\cos^4(x) + \sin^2 (x) + \cos (x)$, when $x \geq 0$?
Any tips on how to solve these types of trigonometric equations would be really helpful. I also tried to solve it with the help of graphs but couldn't manage with 4th degree term of cosine.
 A: Some trig problems can be done with elementary algebra. Let $\cos x=v.$ Then $$\cos^4 x+\sin^2 x+\cos x=\cos^4 x+(1-\cos^2 x)+\cos x=$$ $$=1+v-v^2+v^4=1+v-v^2(1-v^2)\le$$ $$\le 1+v\le 2$$ because $v^2(1-v^2)=(\cos^2x) (1-\sin^2x)\ge 0$ and because $1+v=1+\cos x\le 2.$
Now if $\cos x=1$ then $v=1$ and $1+v-v^2+v^4=2.$
A: If you wanted to use calculus, here is one attempt. Let $p(x)=x^4-x^2+x+1$. In maximising the differentiable function $p(\cos x)$ (a restatement of your original problem), we can find the points of zero derivative - we consider those $x$ such that $p’(\cos x)\cdot\sin x=0$.
One obvious solution is $x=\pi n,n\in\Bbb Z$. It is very straightforward to check that the maximum of $p(\cos x)$ over those $x$ is $p(\cos(2\pi n))=2$.
If we had another solution, where $\sin x\neq0$, then $\cos x$ is a root of $p’(t)=4t^3-2t+1$. Then by basic algebra: $$p(\cos x)=\frac{x}{4}p’(\cos x)-\frac{1}{2}\cos^2x+\frac{3}{4}\cos x+1\le1+\frac{3}{4}\lt2$$So these $x$ do not create maxima since we know there are points with evaluation $2$, so the search is over.
