MHB [ASK]Solution Set of a Trigonometry Inequation

[Monoxdifly]

The set of real numbers x at the interval [0, 2π ] which satisfy $$2sin^2x\geq3cos2x+3$$ takes the form [a, b] ∪ [c, d]. The result of a + b + c + d is ...
a. 4π
b. 5π
c. 6π
d. 7π
e. 8π

What I've done thus far:
$$2sin^2x\geq3cos2x+3$$
$$2sin^2x\geq3(cos2x+1)$$
$$2sin^2x\geq3(cos^2x-sin^2x+sin^2x+cos^2x)$$
$$2sin^2x\geq3(2cos^2x)$$
$$sin^2x\geq3cos^2x$$
$$\frac{sin^2x}{cos^2x}\geq3$$
$$tan^2x\geq3$$
$$tanx\geq\sqrt3$$
$$tanx\geq tan60°$$ or $$tanx\geq tan240°$$
Since the value of tangent become negative after 90° and 270° respectively, I assume that [a, b] ∪ [c, d] is [60°, 90°] ∪ [240°, 270°] thus a + b + c + d = 660° = $$3\frac23\pi$$, but it isn't in the options. Where did I do wrong? I'm sure I solved the inequation alright and just blundered in determining the intervals, but how should I fix them?

 

[skeeter]

dividing by $\cos^2{x}$ was a mistake because $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$ are in the solution set$2\sin^2{x} \ge 3[\cos(2x)+1]$

$2\sin^2{x} \ge 3(1-2\sin^2{x} +1)$

$8\sin^2{x} \ge 6$

$\sin^2{x} - \dfrac{3}{4} \ge 0$

$\left(\sin{x}-\dfrac{\sqrt{3}}{2} \right) \left(\sin{x} + \dfrac{\sqrt{3}}{2} \right) \ge 0$

critical values are $x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3} $

inequality is true for $x$ in the intervals $\left[\dfrac{\pi}{3}, \dfrac{2\pi}{3} \right] \cup \left[\dfrac{4\pi}{3}, \dfrac{5\pi}{3} \right]$

 

[Monoxdifly]

Damn, I should've kept an eye on the denominator. Thanks for your guidance.

 

[I like Serena]

$$tan^2x\geq3$$
$$tanx\geq\sqrt3$$
$$tanx\geq tan60°$$ or $$tanx\geq tan240°$$
Since the value of tangent become negative after 90° and 270° respectively, I assume that [a, b] ∪ [c, d] is [60°, 90°] ∪ [240°, 270°] thus a + b + c + d = 660° = $$3\frac23\pi$$, but it isn't in the options. Where did I do wrong? I'm sure I solved the inequation alright and just blundered in determining the intervals, but how should I fix them?

Also consider that when you take a square root, there is usually a positive and a negative solution.
It should be:
$$\tan^2x\ge 3 \\
|\tan x|\ge \sqrt 3 \\
\tan x \ge \sqrt 3 \quad\lor\quad \tan x \le -\sqrt 3 \\
60° \le x < 90° \quad\lor\quad 90°<x\le 120° \quad\lor\quad 240° \le x < 270° \quad\lor\quad 270°<x\le 300°
$$
Since we accidentally lost the solutions where $\cos x=0$, we can deduce that it is actually:
$$60° \le x \le 120° \quad\lor\quad 240° \le x \le 300° $$

 

[Monoxdifly]

Also consider that when you take a square root, there is usually a positive and a negative solution.
It should be:
$$\tan^2x\ge 3 \\
|\tan x|\ge \sqrt 3 \\
\tan x \ge \sqrt 3 \quad\lor\quad \tan x \le -\sqrt 3 \\
60° \le x < 90° \quad\lor\quad 90°<x\le 120° \quad\lor\quad 240° \le x < 270° \quad\lor\quad 270°<x\le 300°
$$
Since we accidentally lost the solutions where $\cos x=0$, we can deduce that it is actually:
$$60° \le x \le 120° \quad\lor\quad 240° \le x \le 300° $$

That means I'm not completely wrong, am I?