[ASK]Solution Set of a Trigonometry Inequation

[Monox D. I-Fly]

The set of real numbers x at the interval [0, 2π ] which satisfy [tex]2sin^2x\geq3cos2x+3[/tex] 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:

[tex]2sin^2x\geq3cos2x+3[/tex]

[tex]2sin^2x\geq3(cos2x+1)[/tex]

[tex]2sin^2x\geq3(cos^2x-sin^2x+sin^2x+cos^2x)[/tex]

[tex]2sin^2x\geq3(2cos^2x)[/tex]

[tex]sin^2x\geq3cos^2x[/tex]

[tex]\frac{sin^2x}{cos^2x}\geq3[/tex]

[tex]tan^2x\geq3[/tex]

[tex]tanx\geq\sqrt3[/tex]

[tex]tanx\geq tan60°[/tex] or [tex]tanx\geq tan240°[/tex]

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° = [tex]3\frac23\pi[/tex], 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?

[shyamjayakannan]

You omitted a possible case after the [tex]\tan^2x\ge3[/tex] step. Here's the right way to proceed from there:

[tex]\tan^2x\ge3\Rightarrow|\tan x|\ge\sqrt3\Rightarrow\tan x\ge\sqrt3[/tex] and [tex]-\tan x\ge\sqrt3[/tex]. Find the solution for both cases and take the union.