Real roots of the equation $2x^2+(4\sin\theta)x+\cos(2\theta)=0$

Question

For what values of $\theta, 0 ≤ \theta ≤ 2π$ does the equation $$2x^2+(4\sin\theta)x+\cos(2\theta)=0$$ have real roots? Also, am I supposed to find specific values of $\theta$ or find an equation that makes $\theta$ all real roots?

Answer 1

Hint:

As for all quadratic equations, the condition is the (reduced , here) discriminant is non-negative: $$\Delta'=4\sin^2\theta-2\cos 2\theta=2(1-\cos 2\theta)-2\cos 2\theta=2(1-2\cos 2\theta)\ge 0.$$ Can you solve the trigonometric inequation $\cos 2\theta\le\frac12$?

Answer 2

Plug it into the quadratic formula.

$x = 2\sin\theta \pm \sqrt {4\sin^2\theta - 2\cos 2\theta}$

the roots are real if

$4\sin^2\theta - 2\cos 2\theta \ge 0\\ 4\sin^2\theta - 2(1-2\sin^2\theta)\ge 0\\ 8\sin^2\theta - 2\ge 0\\ |\sin\theta| \ge \frac 12\\ \frac {\pi}{6} \le \theta \le \frac {5\pi}{6}$ or
$\frac {5\pi}{6} \le \theta \le \frac {11\pi}{6}$