Answer to Question #147769 in Trigonometry for Moustafa Elemam

Question

(21 - 16〖sin〗^2 x+8 cosx)/(16 〖cos〗^2 x-29-8 √15 sin⁡x ) = 2

find x

Answer 1

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

The Pythagorean Identity

$\sin^2x+\cos^2x=1$

Then

$\dfrac{5+16(1-\sin^2x)+8\cos x}{16(\cos^2x-1)-13-8\sqrt{15}\sin x}=2$

$\dfrac{16\cos^2x+8\cos x+5}{-16\sin^2x-8\sqrt{15}\sin x-13}=2$

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

$\begin{cases} u^2+v^2=1 \\ 16v^2+8v+5=-32u^2-16\sqrt{15}u -26 \end{cases}$

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

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

Then

$\begin{cases} u^2+v^2=1 \\ 4v+1=0 \\ 4u+\sqrt{15}=0 \end{cases}$

Check

$-16\sin^2x-8\sqrt{15}\sin x-13\not=0$

$-16u^2-8\sqrt{15}u-13\not=0$

$u=-\dfrac{\sqrt{15}}{4}:$

$-16(-\dfrac{\sqrt{15}}{4})^2-8\sqrt{15}(-\dfrac{\sqrt{15}}{4})-13$

$=-15+30-13=2\not=0$

Hence

$\cos x=-\dfrac{1}{4}$

$x=\pm(\pi-\arccos(\dfrac{1}{4}))+2\pi n, n\in\Z$