# [ASK]Solution Set of a Trigonometry Inequation

• MHB
• Monoxdifly
In summary, 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 660°.
Monoxdifly
MHB
The set of real numbers x at the interval [0, 2π ] which satisfy $$\displaystyle 2sin^2x\geq3cos2x+3$$ 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:
$$\displaystyle 2sin^2x\geq3cos2x+3$$
$$\displaystyle 2sin^2x\geq3(cos2x+1)$$
$$\displaystyle 2sin^2x\geq3(cos^2x-sin^2x+sin^2x+cos^2x)$$
$$\displaystyle 2sin^2x\geq3(2cos^2x)$$
$$\displaystyle sin^2x\geq3cos^2x$$
$$\displaystyle \frac{sin^2x}{cos^2x}\geq3$$
$$\displaystyle tan^2x\geq3$$
$$\displaystyle tanx\geq\sqrt3$$
$$\displaystyle tanx\geq tan60°$$ or $$\displaystyle tanx\geq tan240°$$
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° = $$\displaystyle 3\frac23\pi$$, 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?

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 $\left[\dfrac{\pi}{3}, \dfrac{2\pi}{3} \right] \cup \left[\dfrac{4\pi}{3}, \dfrac{5\pi}{3} \right]$

Last edited by a moderator:
Damn, I should've kept an eye on the denominator. Thanks for your guidance.

Monoxdifly said:
$$\displaystyle tan^2x\geq3$$
$$\displaystyle tanx\geq\sqrt3$$
$$\displaystyle tanx\geq tan60°$$ or $$\displaystyle tanx\geq tan240°$$
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° = $$\displaystyle 3\frac23\pi$$, 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:
$$60° \le x \le 120° \quad\lor\quad 240° \le x \le 300°$$

Klaas van Aarsen said:
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?

## What is a solution set of a trigonometry inequation?

A solution set of a trigonometry inequation is the set of all possible values of a variable that satisfy the given inequation. In other words, it is the set of values that make the inequation true when substituted for the variable.

## How do you find the solution set of a trigonometry inequation?

To find the solution set of a trigonometry inequation, you need to first isolate the variable on one side of the inequation. Then, use algebraic techniques to solve for the variable. Finally, check your solution by substituting it back into the original inequation to see if it makes the inequation true.

## What are the common types of trigonometry inequations?

The most common types of trigonometry inequations are those involving trigonometric functions such as sine, cosine, and tangent. These inequations can be linear, quadratic, or involve multiple trigonometric functions.

## Why is it important to find the solution set of a trigonometry inequation?

Finding the solution set of a trigonometry inequation is important because it helps us determine the range of values for a variable that satisfies the given inequation. This information is useful in various fields such as engineering, physics, and mathematics.

## Can the solution set of a trigonometry inequation be an infinite set?

Yes, the solution set of a trigonometry inequation can be an infinite set. This can happen when the inequation has no upper or lower bound, or when the variable can take on an infinite number of values that satisfy the inequation.

• General Math
Replies
2
Views
764
• Precalculus Mathematics Homework Help
Replies
1
Views
970
• Differential Equations
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
14
Views
1K
• General Math
Replies
2
Views
6K
• Precalculus Mathematics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Precalculus Mathematics Homework Help
Replies
24
Views
2K
• Precalculus Mathematics Homework Help
Replies
25
Views
2K