Solve the inequality: [tex]\( \sqrt{\sin x} \left( \cos x + \frac{\sqrt{3}}{2} \right) \leq 0 \)[/tex].

To solve the inequality sin x ​ ( cos x + 2 3 ​ ​ ) ≤ 0 , we identify that sin x ​ ≥ 0 and solve for cos x ≤ − 2 3 ​ ​ . The final intervals where both conditions overlap are [ 6 5 π ​ , 6 7 π ​ ] + 2 nπ for integers n .
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Answered by Anonymous | 2025-07-04

To solve the inequality sin x ​ ( cos x + 2 3 ​ ​ ) ≤ 0 , we should consider each factor separately and understand where the inequality could hold true.

Understanding sin x ​ :


The expression sin x ​ is defined only when sin x ≥ 0 , because you cannot have a square root of a negative number.

sin x ≥ 0 is true in the intervals where x = kπ to x = kπ + π , where k is an integer. Specifically, in the range 0 ≤ x ≤ π and repeating every 2 π .



Understanding cos x + 2 3 ​ ​ :


The expression cos x + 2 3 ​ ​ ≤ 0 implies cos x ≤ − 2 3 ​ ​ .

The cosine function is less than or equal to − 2 3 ​ ​ in certain sections of its cycle, specifically between the angles where x = 5 π /6 and x = 7 π /6 within a 0 ≤ x ≤ 2 π interval, and these intervals repeat every 2 π .



Combining the Conditions:


Now we need both conditions to be true simultaneously.

For sin x ≥ 0 and cos x ≤ − 2 3 ​ ​ , this is satisfied on the interval 5 π /6 ≤ x ≤ π on a single cycle.



Conclusion:


Considering the periodic nature of trigonometric functions, the inequality holds for x in the intervals 6 5 π ​ + 2 kπ ≤ x < π + 2 kπ for integer k .

Thus, the solution sets for the inequality occur at certain intervals depending on the periodicity and defined regions for sin x ​ and cos x + 2 3 ​ ​ . These mathematical details ensure the inequality is understood and interpreted correctly.

Answered by DanielJosephParker | 2025-07-06