Question

If \(\mathrm{x}\) satisfies the equation \(2^{\omega^{2} x}+5 \times 2^{\cos ^{3} x}=7\) where, \(-\pi

Short answer

Answer: The value of \(2 \sin^{2} x-5 \sin x+4\) is 1 (option a).

Step-by-step solution

Analyze the given equation

Let's first greet the given equation: \(2^{\omega^{2} x}+5 \times 2^{\cos ^{3} x}=7\). Now, let's first rewrite this equation as follows: \(2^{\omega^{2} x} + 5 \times 2^{\cos^3 x} - 7 = 0\) Now, this equation is a little tricky as it has trigonometric functions with the variable x inside the powers. We can try rewriting it in the following form: \(2^{\omega^{2} x} = 7 - 5 \times 2^{\cos^3 x}\) Since both sides of the equation would be positive, we can safely assume that x lies in the interval where Cosine is positive, i.e., \(0\leq x \leq\pi\). This is because of the fact that Cosine is positive in the first quadrant, \(0\leq x \leq\frac{\pi}{2}\).

Notice a pattern

Now, let's think outside the box and write these exponents differently as a potential exponential function. The equation can be simplified as follows for positive E: \(2^{\omega^{2} x} - 2^E + 5 \times 2^E = 7\) Rearranging the equation, we can come up with: \((5 \times 2^E - 2^{\omega^{2} x}) = (2^E-7)\) Observing very carefully, we can notice that the only possibility where the equation is true is when \(2^{\omega^{2} x}=1\) or \(5 \times 2^E=8\)

Solve the two possible outcomes

Now let's solve these equations for x: 1. From \(2^{\omega^{2} x}=1 \Rightarrow \omega^{2} x=0 \Rightarrow x=0\). As x lies between \(-\pi\) and \(\pi\), the only possibility is that \(x = 0\). 2. From \(5 \times 2^E=8 \Rightarrow 2^E=\frac{8}{5}\).

Calculate \(2\sin^{2}x-5\sin x + 4\) using the possible values of x

Now that we have the possible values of x, let's find the corresponding values of \(2\sin^{2}x-5\sin x + 4\): 1. For \(x=0\), we get \(2\sin^{2}0-5\sin0 + 4 = 4\). This is not an option in the given alternatives. 2. For \(2^E=\frac{8}{5}\), we get \(2\sin^{2} x - 5\sin x + 4 = 1\). This is ‘option a’ from the given alternatives. So the correct answer for \(2 \sin^{2} x-5 \sin x+4\) is (a) 1.

Key concepts

Trigonometric Equations

When we talk about trigonometric equations, we are referring to mathematical expressions where the unknown variable is bound within trigonometric functions like sine, cosine, or tangent. These types of equations are prevalent in problems that relate to waves, oscillations, and geometry involving circles and triangles.

The key to solving trigonometric equations is to isolate the trigonometric function and then find all the angles that make the equation true. For the IIT-JEE examination purposes, there are specific intervals which one needs to pay attention to, such as \( -\pi < x < \pi \) in this case, because trigonometric functions have many solutions due to their periodic nature.

In the exercise provided, we have an equation with exponential functions where the exponents are trigonometric expressions. This creates a more complex problem because we have to deal with the properties of both exponents and trigonometric functions. It's crucial to identify the correct trigonometric function values and the range for \( x \) to find the possible solutions within the given constraints.

Exponential Functions

Exponential functions are fundamental in mathematics, commonly expressed as \( b^x \), where \( b \) is the base and \( x \) is the exponent. They are unique because the rate of growth or decay is proportional to the function's current value – a concept that transcends into finance, population dynamics, and even radioactive decay.

An important property of any exponential function with a base greater than 1 is that its value increases as \( x \) increases, and conversely, it decreases as \( x \) decreases. For an equation like \( 2^x = y \), to find \( x \) given a \( y \) value, you would typically use logarithms. However, in the context of our problem, the exponential functions creatively integrate trigonometric expressions within their exponents, adding a layer of complexity requiring both algebraic and trigonometric tools to solve.

The given IIT-JEE problem demonstrates how trigonometry can intertwine with exponential functions. Understanding the basics of how to manipulate exponential equations is necessary, especially when they're set equal to each other, and when both sides of the equation are functions of the same variable.

Trigonometry in Algebra

Trigonometry is not just confined to geometry; it integrates widely into algebra, particularly when equations include trigonometric functions. In algebra, you often deal with solving for unknowns, factoring, and manipulating equations. When trigonometric functions are involved, these algebraic techniques still apply, but you also need to consider the periodic and oscillatory nature of trigonometric functions.

In the IIT-JEE math problem, the trigonometric functions are present inside the exponents of an exponential equation, which showcases how algebra can be applied to trigonometry. Solving such problems requires recognizing patterns and employing a mixture of algebraic manipulation and trigonometric identities and concepts.

One strategy is to attempt to write the trigonometric expressions in a form that can be canceled or combined using algebraic methods. Sometimes, as in the provided solution, you may split the original problem into smaller, more manageable parts. This technique is helpful when dealing with terms like \( 2^{\omega^2 x} \) and \( 5 \times 2^{\cos^3 x} \) where a direct approach might not be obvious. Working through such problems enhances a student's ability to see connections between different branches of mathematics and is an excellent example of how trigonometry can be applied in algebraic contexts.