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Trigonometric equations like \(|\cos x|^{\sin^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1\) are a fascinating part of Mathematics for JEE Main. These equations involve trigonometric functions and often require careful analysis to solve. When dealing with trigonometric equations, it's essential to recognize that the solution may consist of multiple possibilities. This is because trigonometrical functions, such as sine and cosine, are periodic. Therefore, every solution tends to repeat after a certain interval. In the given exercise,

• The key is to find when the expression \(|\cos x|^{\sin^{2} x- \frac{3}{2} \sin x+\frac{1}{2}}=1\) holds true.
• Recall that any expression of the form \(a^b = 1\) is true if \(a=1\), \(a=-1\) with \(b\) being even, or \(b=0\).

Understanding these basic properties lends significant insight into solving complex trigonometric equations efficiently.

Quadratic equations surface frequently, even in trigonometric contexts. The quadratic equation derived from the given problem is \(y^2 - \frac{3}{2}y + \frac{1}{2} = 0\), where \(y\) represents \(\sin x\).To solve this, we employ the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), for an equation of the form \( ax^2 + bx + c = 0 \). Here,

• \(a = 1\)
• \(b = -\frac{3}{2}\)
• \(c = \frac{1}{2}\)

Using these values, the solutions are \( y = 1 \) and \( y = \frac{1}{2} \). This gives potential values for \( \sin x \), which can then be used to determine \(x\). Understanding quadratic equations not only helps solve this step but also reinforces the concept of how algebraic manipulations can interconnect with trigonometry.

In the realm of trigonometry, functions like sine and cosine are pivotal. These functions are periodic and defined across the unit circle. For cosine, \(\cos x = 1\) means \(x = 2n\pi\), and \(\cos x = -1\) corresponds to \((2n+1)\pi\). On the other hand, for sine, \(\sin x = 1\) occurs at \(x = (2n+1)\frac{\pi}{2}\), while \(\sin x = \frac{1}{2}\) happens at \(x = n\pi + (-1)^n\frac{\pi}{6}\).It’s essential to grasp these core behaviors to find the correct values of \(x\) in trigonometric contexts. By knowing these crucial points within the unit circle, you can efficiently solve trigonometric problems and accurately navigate through the maze that is trigonometric equations.Understanding trigonometrical functions deeply not only helps in mathematics but also enhances a learner's problem-solving abilities for exams like JEE Main.