You will surely learn something from it!

$c={ lim }_{ x\rightarrow \cfrac { \pi }{ 2 } }(a{ \quad cos }^{ 4 }x+b{ \quad cos }^{ 2 }x+c)\\ \\ \quad ={ lim }_{ x\rightarrow \cfrac { \pi }{ 2 } }\cfrac { cos5x }{ cosx } \quad (\cfrac { 0 }{ 0 } \quad form)\\ \\ \quad ={ lim }_{ x\rightarrow \cfrac { \pi }{ 2 } }\cfrac { -5sin5x }{ -sinx } =5$

I used $\theta$ instead of $x$ . Anyways, no calculus required!

Note that $\cos5\theta$ is simply the real part of $e^{5i\theta}=\cos5\theta + i\sin5\theta$ .

By DeMoivre's Theorem , We have $\cos5\theta+i\sin 5\theta=(cos\theta+i\sin\theta)^5$ .

Expanding, and taking note of the identity $\sin^2 \theta=1- \cos^2 \theta$ , the real part is then $16\cos^5 \theta -20\cos^3 \theta+5\cos\theta$ .

Therefore, we have $c=\boxed{5}$ .

$\large\cos 5x=a\cos^5 x+b\cos^3 x+c\cos x$ Differentiating both sides w.r.t $x$ , $\large\Rightarrow -5\sin 5x=-5a \cos^4x•\sin x-3b\cos^2x•\sin x-c\sin x$ Here we observe that $\color{#3D99F6}{\text {every terms of RHS has sin x with it except the last term containing c.}}$ We can take advantage of this fact by, putting $x=\frac{π}{2}$

We finally get, $\large\Rightarrow -5 \sin(\frac{5π}{2})=-c(how)$ $\Large\Rightarrow c=\boxed5$

$Cos5X=Cos(4X+X)=Cos4X*CosX - Sin4X*SinX\\=(2Cos^22X-1)*CosX-2Sin2X*Cos2x*SinX\\=(2Cos^22X-1)*CosX-4Sin^2X*CosX*Cos2X~~~~~ ~~~ \\ =CosX\{2Cos^22X-1-4(1-Cos^2X)*Cos2X \} \\=CosX\{2(2Cos^2X -1)^2-1-4(1-Cos^2X)*(2Cos^2X -1)\\ =CosX\{2(2Cos^2X -1)^2-1-4(1-Cos^2X)*(2Cos^2X -1)\\ =CosX\{2(4Cos^4X - 4Cos^2X +1) -1-4( - 1+3Cos^2X - 2Cos^4X \}\\ =CosX\{2-1+4~ plus~ terms~ containing~ CosX\} ~~\implies ~c=5.$