# Question 3aca5

Jun 16, 2017

(7pi)/6; (11pi)/6#

#### Explanation:

1/(sec^2 x) + 3sin (x/2)cos (x/2) = 0
since $\frac{1}{{\sec}^{2} x} = {\cos}^{2} x = 1 - {\sin}^{2} x$, and
$3 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) = \left(\frac{3}{2}\right) \sin x$, therefore:
$1 - {\sin}^{2} x + \left(\frac{3}{2}\right) \sin x = 0$
$- 2 {\sin}^{2} x + 3 \sin x + 2 = 0$
Solve this quadratic equation for sin x.
$D = {d}^{2} = {b}^{2} - 4 a c = 9 + 16 = 25$ --> $d = \pm 5$
There are 2 real roots:
$\sin x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{3}{-} 4 \pm \frac{5}{4} = \frac{3}{4} \pm \frac{5}{4}$
$\sin x = \frac{8}{4} = 2$ (rejected as > 1), and
$\sin x = - \frac{2}{4} = - \frac{1}{2}$
Trig table and unit circle give:
$x = \frac{- \pi}{6}$ or $x = \frac{11 \pi}{6}$ (co-terminal), and
$x = \pi - \left(- \frac{\pi}{6}\right) = x + \frac{\pi}{6} = \frac{7 \pi}{6}$
Check by calculator.
$x = \frac{7 \pi}{6} = {210}^{\circ}$ --> cos x = -0.866 --> ${\cos}^{2} x = 0.75$ -->
x/2 = 105 --> cos (x/2) = -0.258 --> sin (x/2) = 0.966.
cos^2 x + 3sin (x/2).cos (x/2) = 0.75 + 3(-0.258)(0.966) =
= 0.75 - 0.75 = 0. Correct.