Question

Let 21 - 16sin^2(x) + 8cosx / 16cos^2(x) - 29 - 8√15sinx = 2. Find the largest value of 7cosx.

Answer 1

$$\dfrac{21-16\sin^2(x)+8\cos(x)}{16\cos^2(x)-29-8\sqrt{15}\sin(x)}=2$$

Let $v=\cos(x), u=\sin(x).$ Then

$$u^2+v^2=1$$

$$16v^2-29-8\sqrt{15}u\not=0$$

$$21-16u^2+8v-32v^2+58+16\sqrt{15}u=0$$

$$(16v^2+8v+1)+2(16u^2+8\sqrt{15}u+15)=0$$

$$(4v+1)^2+2(4u+\sqrt{15})^2=0$$

Then

$$\begin{cases} 4v+1=0 \\ 4u+\sqrt{15}=0 \end{cases}=>\begin{cases} v=-\dfrac{1}{4} \\ u=-\dfrac{\sqrt{15}}{4} \end{cases}$$

Check

$$u^2+v^2=(-\dfrac{1}{4})^2+(-\dfrac{\sqrt{15}}{4})^2=1, True$$

$$16v^2-29-8\sqrt{15}u=$$

$$=16(-\dfrac{1}{4})^2-29-8\sqrt{15}(-\dfrac{\sqrt{15}}{4})=1-29+30=$$

$$=2\not=0, True$$

Hence

$\cos(x)=v=-\dfrac{1}{4}$

$7\cos(x)=-\dfrac{7}{4}=-1.75$