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Chapter 4: Problem 8

Solution of equation \(\tan ^{-1} 2 x+\tan ^{-1} 3 x=3 \pi / 4\) is (a) \(x=1,-\frac{1}{6}\) (b) \(x=-\frac{1}{6}\) (c) \(\mathrm{x}=1 / 6,-1\) (d) 1

Short Answer

Expert verified
The solution to the equation \(\tan^{-1}2x+\tan^{-1}3x = \frac{3\pi}{4}\) using the given steps results in \(x=\frac{1}{2}\) and \(x=\frac{1}{3}\), which do not match any of the provided answer choices. There is likely an error in the problem statement or the answer choices.

Step by step solution

01

Apply the sum formula for inverse tangent

We are given the equation \(\tan^{-1} 2x + \tan^{-1} 3x = \frac{3\pi}{4}\). Using the sum formula for inverse tangent, we can rewrite this equation as: \(\tan^{-1}\left(\frac{2x+3x}{1-6x^{2}}\right)=\frac{3 \pi}{4}\)
02

Take tangent of both sides

Taking the tangent of both sides, we obtain: \(\frac{2x+3x}{1-6x^{2}}=\tan \left(\frac{3 \pi}{4}\right)\) Since \(\tan(\frac{3\pi}{4}) = -1\), we have: \(\frac{2x+3x}{1-6x^2} = -1\)
03

Solve for x

Now, we will solve for x: \(\frac{5x}{1-6x^2} = -1\) Multiply both sides with \(1-6x^2\): \(5x = -(1-6x^2)\) Expand and rearrange the equation: \(5x + 1 = 6x^2\) Move all the terms to one side: \(6x^2 - 5x + 1 = 0\) Now, we can solve this quadratic equation by finding its roots. The quadratic equation is already in the proper form (\(ax^2+bx+c=0\)). We can proceed to use the quadratic formula: \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
04

Apply the quadratic formula

Applying the quadratic formula, we have: \(x=\frac{-(-5)\pm\sqrt{(-5)^2-4(6)(1)}}{2(6)}=\frac{5\pm\sqrt{25-24}}{12}\) Now, we can simplify the expression: \(x=\frac{5\pm\sqrt{1}}{12}\) There are now two options for x: 1. If we choose the positive square root: \(x = \frac{5+1}{12} = \frac{6}{12} = \frac{1}{2}\) 2. If we choose the negative square root: \(x = \frac{5-1}{12} = \frac{4}{12} = \frac{1}{3}\) However, neither of these options match the given answers. Therefore, it's likely that an error has been made in the problem statement or the answer choices.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Equations
Trigonometric equations involve trigonometric functions such as sine, cosine, and tangent, and require specific methods to solve. These equations might look daunting at first, but there are key strategies that can help simplify the problem.

For example, to solve an equation with multiple inverse tangent functions, like in our exercise, we can apply well-known trigonometric identities, such as the inverse tangent sum formula. This allows us to combine the tangent expressions into a single inverse tangent function. The key is to transform the equation into a form that can be easily managed and, ultimately, solved.

In some cases, after applying identities, we might end up with algebraic equations that can be simplified further. It's essential to check the initial conditions after finding a solution, as the range of inverse trigonometric functions is limited, which might influence the final answer set.
Quadratic Formula
The quadratic formula is an essential tool for solving quadratic equations, which are of the form \(ax^2 + bx + c = 0\). The formula provides a systematic way to find the roots of any quadratic equation by considering the coefficients \(a\), \(b\), and \(c\).

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). To use it effectively, first ensure that the quadratic equation is in the standard form. Then, identify the coefficients and plug them into the formula. It's important to consider both the positive and negative solutions given by the \(\pm\) symbol, as they can both be valid solutions to the problem at hand.

Solving quadratic equations is straightforward with the quadratic formula, but always simplify your answers. Sometimes, mistakes in simplification might lead to incorrect solutions, as seen in the original exercise solution where the simplification led to answers that didn't match any of the provided options.
Inverse Trigonometric Functions
Inverse trigonometric functions, often denoted with a raised \(-1\) notation (like \(\tan^{-1}\)), allow us to find the angles when the value of a trigonometric function is known. Each inverse trigonometric function has a specific domain and range, which dictate the possible values of angles that can be returned.

For instance, the inverse tangent function, \(\tan^{-1}\), will always return an angle in the \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) range. This property is vital because it affects the set of solutions we can consider for an equation involving inverse trigonometric functions.

When working with these functions, it's important to consider the quadrants in which the angle lies, based on the sign of the trigonometric value, and also to be aware of principal values, which are the most commonly used values within the restricted domain. Understanding these functions and their properties can significantly facilitate the process of solving trigonometric equations.

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Most popular questions from this chapter

\(f(k)-100<0\) \(\Rightarrow k^{2}-100<0\) \(\Rightarrow k \in(-10,10)\) But \(k \in \mathbb{N}\) Solution set of given inequality will be \(\\{1,2,3,4,5, \ldots 9\\}\)

The questions given below consist of an assersion (A) and the reason (R). Use the following key to choose the appropriate answer. (a) If both assertion and reason are correct and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct but reason is not correct explanation of the assertion. (c) If assertion is correct, but reason is incorrect (d) If assertion is incorrect, but reason is correct Now consider the following statements: A: If \(x<0, \tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}\) R: \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}, \forall x \in R\).

The sum of infinite terms of the series \(\tan ^{-1}\left(\frac{2}{1-1^{2}+1^{4}}\right)+\tan ^{-1}\left(\frac{4}{1-2^{2}+2^{4}}\right)+\) \(\tan ^{-1}\left(\frac{6}{1-3^{2}+3^{4}}\right)+\ldots .\) is (a) \(\frac{\pi}{4}\) (b) \(\frac{\pi}{2}\) (c) \(\frac{3 \pi}{4}\) (d) None of these

If two angles of a \(\Delta\) are \(\tan ^{-1} \frac{1}{2}\) and \(\tan ^{-1} 1 / 3\), then find \(\frac{\theta}{15}\) (where \(\theta\) is third angle of \(\Delta\) in degree).

The least and greatest values of \(\left(\sin ^{-1} x\right)^{3}+\left(\cos ^{-1} x\right)^{3}\) are respectively (a) \(-\pi / 2, \pi / 2\) (b) \(-\pi^{3} / 8, \pi^{3} / 8\) (c) \(\pi^{3} / 32,7 \pi^{3} / 8\) (d) None of these
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