Problem 11 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(0 \leq x<\frac{\pi}{2}\), then the solution of the equation \(16^{\sin ^{2} x}+16^{\cos ^{2} x}=10\) is given by \(x\) equal to (A) \(\frac{\pi}{6}, \frac{\pi}{3}\) (B) \(\frac{\pi}{3}, \frac{\pi}{2}\) (C) \(\frac{\pi}{6}, \frac{\pi}{2}\) (D) None of these

Step-by-Step Solution

Answer

Option (A) \(\frac{\pi}{6}, \frac{\pi}{3}\) is correct.

Step 1: Understand the Equation

We are given the equation \(16^{ ext{sin}^2 x} + 16^{ ext{cos}^2 x} = 10\) and the range \(0 \leq x < \frac{\pi}{2}\). Our goal is to find values of \(x\) that satisfy this equation.

Step 2: Simplify the Equation with a Substitution

Let's use the substitution \(a = \sin^2 x\) and \(b = \cos^2 x\), keeping in mind that \(a + b = 1\). Rewriting the equation, we have: \(16^a + 16^b = 10\).

Step 3: Explore Values for Simplification

Since \(a + b = 1\), we suspect either \(a\) or \(b\) must be simple fractions such as \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{2}{3}\). Try \(a = \frac{1}{3}\) and \(b = \frac{2}{3}\), and calculate \(16^a\) and \(16^b\).

Step 4: Verify Each Solution

With \(a = \frac{1}{3}\), we find \(16^{1/3} + 16^{2/3} = 10\). This is correct because \(2 + 8 = 10\). Thus, we need to find \(x\) values such that \(\sin^2 x = \frac{1}{3}\) or \(\cos^2 x = \frac{1}{3}\).

Step 5: Calculate Possible Values for \(x\)

Solve \(\sin x = \pm \sqrt{\frac{1}{3}}\) and \(\cos x = \pm \sqrt{\frac{1}{3}}\). Find corresponding angles within \(0 \leq x < \frac{\pi}{2}\): \(x = \frac{\pi}{6}\) or \(x = \frac{\pi}{3}\). Both values satisfy the original equation.

Step 6: Conclusion

Thus the solutions must correspond to option (A), where \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{3}\).

Key Concepts

Substitution method in trigonometry

The substitution method in trigonometry helps solve equations by replacing complex expressions with simpler variables. In this particular problem, we begin by substituting \(a = \sin^2 x\) and \(b = \cos^2 x\). This substitution simplifies the trigonometric expressions into more manageable forms.

Here, for example, we utilize the identity \(a + b = 1\), which arises from the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). With this substitution, we transform our original equation \(16^{\sin^2 x} + 16^{\cos^2 x} = 10\) into \(16^a + 16^b = 10\).

This transformation allows us to focus on finding values for \(a\) and \(b\) that are easier to handle, shifting from trigonometric functions to simple algebraic expressions. This approach is especially useful when combined with trigonometric identities, providing a pathway to solve for variables without getting tangled in complex trigonometric behavior.

Solutions of trigonometric equations

Finding the solutions of trigonometric equations involves determining the values of variables that make equations true. For our equation \(16^{\sin^2 x} + 16^{\cos^2 x} = 10\), after substitution and simplification, the problem focuses on the equation \(16^a + 16^b = 10\) where \(a + b = 1\).

We're solving for \(a\) and \(b\) keeping in mind constraints of \(a = \sin^2 x\) and \(b = \cos^2 x\). Given \(a + b = 1\) ensures that we are still working within the unit circle of trigonometric identities. By evaluating potential values like \(a = \frac{1}{3}\) and \(b = \frac{2}{3}\), we test simple fractions because they often result in neat, exact numbers that satisfy the equation.

Once solutions for \(a\) and \(b\) satisfy the algebraic equation, they must be translated back into trigonometric results. We look for \(x\) such that these fractions occur within the specified interval \(0 \leq x < \frac{\pi}{2}\). This process shows that \(x\) values \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\) are the desired solutions, as they satisfy both the simplified algebraic equation and return to meaningful trigonometric values.

Trigonometric identities

Trigonometric identities are pivotal in transforming and solving equations by revealing relationships between trigonometric functions. In this problem, the identity \(\sin^2 x + \cos^2 x = 1\) serves as the core principle that simplifies the given equation.

This identity converts our initial expression into a familiar form, suitable for algebraic manipulation. It ensures that we can directly relate the squares of sine and cosine to unity, providing a deterministic approach to identify solutions. This identity also reassures us that if \(\sin^2 x\) is known, \(\cos^2 x\) is instantly determined, and vice versa.

Moreover, these identities help in checking the reasonableness of solutions post-substitution by ensuring that calculated values fall within the expected range of trigonometric functions. Once the transformation using trigonometric identities is performed, the equation becomes straightforward to solve, ultimately leading to correct answers like \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{3}\), confirming the solutions correspond to the options given in the exercise.