Question b59e0

Sep 6, 2017

$\arccos \frac{\pm \sqrt{\frac{3}{5}}}{2}$

Explanation:

Let $X = \cos \left(2 x\right)$ and $Y = \sin \left(2 x\right)$, the condition i equivalent to
$11 + {Y}^{2} / {X}^{2} = \frac{7}{X} ^ 2 \implies 11 {X}^{2} + {Y}^{2} = 7$ and ${X}^{2} + {Y}^{2} = 1$
Subtracting the latest 2 condition we get $10 {X}^{2} = 6$
So ${X}^{2} = \frac{3}{5} \implies X = \pm \sqrt{\frac{3}{5}}$
$\cos \left(2 x\right) = \pm \sqrt{\frac{3}{5}}$

Sep 8, 2017

30^@; 39^@23#

Explanation:

$11 + \frac{{\sin}^{2} 2 x}{{\cos}^{2} 2 x} = \frac{7}{\cos 2 x}$
$11 \cos 2 x + \frac{{\sin}^{2} 2 x}{\cos 2 x} = 7$
Multiply both sides by cos 2x
$11 {\cos}^{2} 2 x + {\sin}^{2} 2 x = 7 \cos 2 x$
Replace ${\sin}^{2} 2 x$ by $\left(1 - {\cos}^{2} 2 x\right)$
$11 {\cos}^{2} 2 x + 1 - {\cos}^{2} 2 x = 7 \cos 2 x$
Solve this quadratic equation for cos 2x
$10 {\cos}^{2} 2 x - 7 \cos 2 x + 1 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 49 - 40 = 9$ --> $d = \pm 3$
There are 2 real roots:
$\cos 2 x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{7}{20} \pm \frac{3}{20}$
$\cos 2 x = \frac{10}{20} = \frac{1}{2}$ and $\cos 2 x = \frac{4}{20} = \frac{1}{5}$

a. $\cos 2 x = \frac{1}{2}$ --> $2 x = \pm \frac{\pi}{3}$,
$x = \pm \frac{\pi}{6}$, or $x = \pm {30}^{\circ}$
b. $\cos 2 x = \frac{1}{5}$ --> $2 x = \pm {78}^{\circ} 46$ -->
$x = \pm {39}^{\circ} 23$
$x = {30}^{\circ}$ and $x = {39}^{\circ} 23$