Problem 221 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left\\{\begin{array}{l} \cos x \leq \frac{1}{\sqrt{2}} \\ \cot x>-\sqrt{3} \end{array}\right\\} $$

Step-by-Step Solution

Answer

The solution to the system of inequalities is when \(x\) is in the range \([0, \frac{\pi}{4}]\) or \([\frac{3\pi}{4}, \pi]\) or \([2n\pi, 2n\pi + \frac{\pi}{4}]\) or \([2n\pi + \frac{3\pi}{4}, 2n\pi + \pi]\) where \(n\) is an integer, intersected with the range \((-\frac{\pi}{2}, -\frac{\pi}{3})\) or \((\frac{\pi}{3}, \frac{\pi}{2})\) or \((2n\pi - \frac{\pi}{2}, 2n\pi - \frac{\pi}{3})\) or \((2n\pi + \frac{\pi}{3}, 2n\pi + \frac{\pi}{2})\) where \(n\) is an integer.

Step 1: Solve the first inequality

The inequality \(\cos x \leq \frac{1}{\sqrt{2}}\) needs to be solved. Remember that \(\cos x\) is less than or equal to \(\frac{1}{\sqrt{2}}\) when \(x\) is in the range \([0, \frac{\pi}{4}]\) or \([\frac{3\pi}{4}, \pi]\) or \([2n\pi, 2n\pi + \frac{\pi}{4}]\) or \([2n\pi + \frac{3\pi}{4}, 2n\pi + \pi]\) where \(n\) is an integer.

Step 2: Solve the second inequality

The second inequality \(\cot x > -\sqrt{3}\) needs to be solved. First, change cotangent to tangent using the relationship \(\cot x = \frac{1}{\tan x}\). Thus the inequality becomes \(-\tan x < \sqrt{3}\) which is true for \(x\) in the range \((-\frac{\pi}{2}, -\frac{\pi}{3})\) or \((\frac{\pi}{3}, \frac{\pi}{2})\) or \((2n\pi - \frac{\pi}{2}, 2n\pi - \frac{\pi}{3})\) or \((2n\pi + \frac{\pi}{3}, 2n\pi + \frac{\pi}{2})\) where \(n\) is an integer.

Step 3: Find the intersection of the solution sets

The solution to the systems of inequalities is the intersection of the individual solution sets from Step 1 and Step 2.

Key Concepts

Cosine Inequality

Understanding trigonometric inequalities often starts with the specific functions themselves. The cosine function, denoted as \(\cos x\), is crucial when dealing with these types of inequalities. Here, we are given the inequality \(\cos x \leq \frac{1}{\sqrt{2}}\). To solve this, we need to consider the behavior of the cosine function.

• The cosine function is positive in the first and fourth quadrants of the unit circle.
• Since its maximum value is 1, \(\cos x\) reaches \(\frac{1}{\sqrt{2}}\) at \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).
• The solution extends from \([0, \frac{\pi}{4}])\) and \([\frac{3\pi}{4}, \pi])\), continuing in intervals due to the periodicity of the cosine function.

Understanding these ranges helps identify the complete solution. Remember, the cosine function's periodic nature means we consider patterns that repeat every \(2\pi\), hence the expression \(2n\pi\) is involved with integer \(n\). This is key when generalizing the solution across all possible cycles.

Cotangent Inequality

The cotangent function, represented as \(\cot x\), offers its own set of challenges in inequality solutions. In our problem, we examine the inequality \(\cot x > -\sqrt{3}\). Often, transforming cotangent to a tangent function can simplify the process because tangent is generally more intuitive to work with.

• The relationship between cotangent and tangent is \(\cot x = \frac{1}{\tan x}\), which means that \(\cot x > -\sqrt{3}\) transforms into \(-\tan x < \sqrt{3}\).
• This inequality bestows us with a new angle range primarily in the second and fourth quadrants of the unit circle.
• Consequently, solutions are found within the intervals \((\frac{\pi}{3}, \frac{\pi}{2})\) and \((-\frac{\pi}{2}, -\frac{\pi}{3})\), extended periodically as \((2n\pi + \frac{\pi}{3}, 2n\pi + \frac{\pi}{2})\).

For cotangent or any reciprocal trigonometric function, always verify the behavior of the function over various intervals to understand where it meets the inequality conditions.

Solution of Inequalities

Solving trigonometric inequalities generally requires finding the intersection of solutions. For our set of inequalities, it means capturing where both conditions are simultaneously true.

• Solutions for \(\cos x \leq \frac{1}{\sqrt{2}}\) and \(\cot x > -\sqrt{3}\) are determined separately first.
• Next, we find where these two sets overlap using their respective ranges. The overlap points lie within each of their defined sections, and these intersections are typically found where ranges from each step coincide.
• Determining intersections often involves visualizing on the unit circle or using periodicity properties to cross-reference intervals.

When solving, patience is required, as manipulating the expressions helps better visualize where functions "work together." This skill is invaluable for effectively solving any trigonometric inequality, especially when multiple levels of conditions are involved.