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JEE/EAMCET/BITSAT Math: How to Solve Challenging Trigonometric Inequality | 2^{\sec^2 x} \sqrt{\tan^2 y - \tan y + \frac{1}{2}} \le 1
✍️ YouTube Video Description:
Struggling with complex trigonometric inequalities? This video breaks down a frequently asked, high-level question that tests your understanding of both trigonometric function ranges and quadratic minimum values!
The Question:
Find the number of values of x in the interval x \in [0, 2\pi] that satisfy the inequality:
2^{\sec^2 x} \sqrt{\tan^2 y - \tan y + \frac{1}{2}} \le 1
🎯 What you will learn in this video:
1. Finding the Minimum Value of a Quadratic: We show you how to complete the square on the \sqrt{\tan^2 y - \tan y + \frac{1}{2}} term to find its absolute minimum value.
2. Using Trigonometric Ranges: We review the range of \sec^2 x and use it to determine the minimum value of the exponential term 2^{\sec^2 x}.
3. Solving the Inequality: The key insight! By determining the minimum possible value of the entire Left-Hand Side (LHS) of the inequality, we show that the LHS can only satisfy the condition \le 1 if it is exactly equal to 1.
4. Final Solution: We solve for x when 2^{\sec^2 x} is at its minimum, leading to the final set of solutions for x \in [0, 2\pi].
This is a must-watch for students preparing for JEE Main, JEE Advanced, BITSAT, EAMCET, or any competitive math exam where trigonometric functions are covered. Master this concept to score high in Trigonometry!
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