Consider the following values of x: 1) 8 2) -4 3) \(\frac 16\) 4) \(- \frac{1}{4}\) Which of the above values of x is/are the solution(s) of the equation \({\tan ^{ - 1}}\left( {2x} \right) + {\tan ^{ - 1}}\left( {3x} \right) = \frac{\pi }{4}?{\rm{\;}}\)
3 only
Solving the Trigonometric Equation \({\tan ^{ - 1}}\left( {2x} \right) + {\tan ^{ - 1}}\left( {3x} \right) = \frac{\pi }{4}\)
We are asked to find which of the given values of \(x\) are solutions to the equation \({\tan ^{ - 1}}\left( {2x} \right) + {\tan ^{ - 1}}\left( {3x} \right) = \frac{\pi }{4}\).
To solve this trigonometric equation, we can use the formula for the sum of inverse tangents: \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B\). The formula has different forms depending on the product \(AB\).
The general formula is:
- \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\) if \(AB < 1\)
- \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = \pi + {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\) if \(AB > 1\) and \(A > 0, B > 0\)
- \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = -\pi + {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\) if \(AB > 1\) and \(A < 0, B < 0\)
- \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = \frac{\pi}{2}\) if \(AB = 1\) and \(A > 0, B > 0\)
- \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = -\frac{\pi}{2}\) if \(AB = 1\) and \(A < 0, B < 0\)
In our equation, \(A = 2x\) and \(B = 3x\). So the product \(AB = (2x)(3x) = 6x^2\).
Let's assume that \(6x^2 < 1\). Using the first formula, we get:
\[{\tan ^{ - 1}}\left( {\frac{{2x + 3x}}{{1 - 6x^2}}} \right) = \frac{\pi }{4}\] \[{\tan ^{ - 1}}\left( {\frac{{5x}}{{1 - 6x^2}}} \right) = \frac{\pi }{4}\]Taking the tangent of both sides:
\[\frac{{5x}}{{1 - 6x^2}} = \tan\left(\frac{\pi}{4}\right)\] \[\frac{{5x}}{{1 - 6x^2}} = 1\]Assuming \(1 - 6x^2 \neq 0\), we can cross-multiply:
\[5x = 1 - 6x^2\]Rearranging the terms, we get a quadratic equation:
\[6x^2 + 5x - 1 = 0\]We can solve this quadratic equation by factoring. We look for two numbers that multiply to \(6 \times -1 = -6\) and add up to 5. These numbers are 6 and -1.
\[6x^2 + 6x - x - 1 = 0\] \[6x(x + 1) - 1(x + 1) = 0\] \[(6x - 1)(x + 1) = 0\]This gives us two potential solutions for \(x\):
\[6x - 1 = 0 \implies 6x = 1 \implies x = \frac{1}{6}\] \[x + 1 = 0 \implies x = -1\]Now, we must check if these potential solutions satisfy the condition \(6x^2 < 1\) that we assumed to use the formula \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\).
- For \(x = \frac{1}{6}\): Calculate \(6x^2 = 6\left(\frac{1}{6}\right)^2 = 6\left(\frac{1}{36}\right) = \frac{6}{36} = \frac{1}{6}\). Since \(\frac{1}{6} < 1\), the condition \(6x^2 < 1\) is satisfied. Thus, \(x = \frac{1}{6}\) is a valid solution derived under this condition.
- For \(x = -1\): Calculate \(6x^2 = 6(-1)^2 = 6(1) = 6\). Since \(6 \not< 1\), the condition \(6x^2 < 1\) is not satisfied. For \(x=-1\), \(6x^2 = 6 > 1\). Also, for \(x=-1\), \(2x = -2\) and \(3x = -3\), both negative. In this case, the correct formula is \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = -\pi + {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\). Let's check the LHS for \(x=-1\): \begin{align*} {\tan ^{ - 1}}(-2) + {\tan ^{ - 1}}(-3) &= -\pi + {\tan ^{ - 1}}\left( {\frac{{-2 + (-3)}}{{1 - (-2)(-3)}}} \right) \\ &= -\pi + {\tan ^{ - 1}}\left( {\frac{{-5}}{{1 - 6}}} \right) \\ &= -\pi + {\tan ^{ - 1}}\left( {\frac{-5}{-5}} \right) \\ &= -\pi + {\tan ^{ - 1}}(1) \\ &= -\pi + \frac{\pi}{4} \\ &= -\frac{3\pi}{4}\end{align*} Since \(-\frac{3\pi}{4} \neq \frac{\pi}{4}\), \(x = -1\) is not a solution to the original equation.
So, from the solutions to the quadratic equation, only \(x = \frac{1}{6}\) is a valid solution to the original trigonometric equation.
Now let's check the given values of \(x\) from the options:
- \(x = 8\): \(2x = 16\), \(3x = 24\). \(6x^2 = 6(8)^2 = 6(64) = 384\). Since \(384 > 1\) and \(2x, 3x\) are positive, the formula is \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = \pi + {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\). The LHS is \(\pi + {\tan ^{ - 1}}\left( {\frac{{16 + 24}}{{1 - 384}}} \right) = \pi + {\tan ^{ - 1}}\left( {\frac{40}{-383}} \right)\). This value is in the range \((\frac{\pi}{2}, \pi)\), which is not equal to \(\frac{\pi}{4}\). So \(x=8\) is not a solution.
- \(x = -4\): \(2x = -8\), \(3x = -12\). \(6x^2 = 6(-4)^2 = 6(16) = 96\). Since \(96 > 1\) and \(2x, 3x\) are negative, the formula is \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = -\pi + {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\). The LHS is \(-\pi + {\tan ^{ - 1}}\left( {\frac{{-8 + (-12)}}{{1 - 96}}} \right) = -\pi + {\tan ^{ - 1}}\left( {\frac{-20}{-95}} \right) = -\pi + {\tan ^{ - 1}}\left( {\frac{4}{19}} \right)\). This value is in the range \((-\pi, -\frac{\pi}{2})\), which is not equal to \(\frac{\pi}{4}\). So \(x=-4\) is not a solution.
- \(x = \frac{1}{6}\): We already checked this value. \(2x = \frac{1}{3}\), \(3x = \frac{1}{2}\). \(6x^2 = \frac{1}{6} < 1\). The LHS is \({\tan ^{ - 1}}\left(\frac{1}{3}\right) + {\tan ^{ - 1}}\left(\frac{1}{2}\right) = {\tan ^{ - 1}}\left( {\frac{{1/3 + 1/2}}{{1 - (1/3)(1/2)}}} \right) = {\tan ^{ - 1}}\left( {\frac{{5/6}}{{5/6}}} \right) = {\tan ^{ - 1}}(1) = \frac{\pi}{4}\). This matches the RHS. So \(x = \frac{1}{6}\) is a solution.
- \(x = -\frac{1}{4}\): \(2x = -\frac{1}{2}\), \(3x = -\frac{3}{4}\). \(6x^2 = 6(-\frac{1}{4})^2 = 6(\frac{1}{16}) = \frac{6}{16} = \frac{3}{8}\). Since \(\frac{3}{8} < 1\), we might use the formula \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)\). The LHS is \({\tan ^{ - 1}}\left(-\frac{1}{2}\right) + {\tan ^{ - 1}}\left(-\frac{3}{4}\right)\). Since \(2x\) and \(3x\) are negative, both \({\tan ^{ - 1}}\left(-\frac{1}{2}\right)\) and \({\tan ^{ - 1}}\left(-\frac{3}{4}\right)\) are in the range \((-\frac{\pi}{2}, 0)\). Their sum must be in the range \((-\pi, 0)\). The RHS of the original equation is \(\frac{\pi}{4}\), which is positive. A value in \((-\pi, 0)\) cannot equal \(\frac{\pi}{4}\). Therefore, \(x = -\frac{1}{4}\) is not a solution. Alternatively, using the formula: \({\tan ^{ - 1}}\left( {\frac{{-1/2 + (-3/4)}}{{1 - (-1/2)(-3/4)}}} \right) = {\tan ^{ - 1}}\left( {\frac{{-5/4}}{{1 - 3/8}}} \right) = {\tan ^{ - 1}}\left( {\frac{{-5/4}}{{5/8}}} \right) = {\tan ^{ - 1}}(-2)\). This is in \((-\frac{\pi}{2}, 0)\) and not equal to \(\frac{\pi}{4}\). So \(x = -\frac{1}{4}\) is not a solution.
Based on our checks, only the value \(x = \frac{1}{6}\) is a solution to the given equation.
Let's review the options provided:
| Option | Value(s) of x | Is it/Are they solution(s)? |
|---|---|---|
| 1 | 3 only (\(x = \frac{1}{6}\)) | Yes, \(x = \frac{1}{6}\) is a solution. |
| 2 | 2 and 3 only (\(x = -4\) and \(x = \frac{1}{6}\)) | No, \(x = -4\) is not a solution. |
| 3 | 1 and 4 only (\(x = 8\) and \(x = -\frac{1}{4}\)) | No, neither \(x = 8\) nor \(x = -\frac{1}{4}\) are solutions. |
| 4 | 4 only (\(x = -\frac{1}{4}\)) | No, \(x = -\frac{1}{4}\) is not a solution. |
The only value from the options that is a solution is \(x = \frac{1}{6}\), which corresponds to option 1 (3 only).
Revision Table: Key Concepts for Solving Trigonometric Equations
| Concept | Description | Relevance to Problem |
|---|---|---|
| Inverse Tangent Function (arctan or \({\tan ^{ - 1}}\)) | The inverse function of tangent. \(y = {\tan ^{ - 1}}x\) means \(\tan y = x\), where \(y \in (-\frac{\pi}{2}, \frac{\pi}{2})\). | The equation involves inverse tangent functions. Understanding their domain and range is crucial. |
| Sum of Inverse Tangents Formula | Formulae for \({\tan ^{ - 1}}A + {\tan ^{ - 1}}B\), dependent on the value of \(AB\). | This formula is the primary tool used to simplify the LHS of the given equation. Choosing the correct form is essential. |
| Quadratic Equation | An equation of the form \(ax^2 + bx + c = 0\). Solutions can be found by factoring or the quadratic formula. | Simplifying the trigonometric equation led to a quadratic equation in terms of \(x\). |
| Extraneous Solutions | Solutions that arise during the solving process (e.g., from squaring both sides or using conditional formulas) but do not satisfy the original equation. | We must check potential solutions obtained from the quadratic equation against the conditions under which the trigonometric identity was applied. |
Additional Information on Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse functions of the trigonometric functions (sine, cosine, tangent, cosecant, secant, cotangent). They are used to find the angle when the value of the trigonometric function is known. For example, \({\sin ^{ - 1}}(0.5)\) gives the angle whose sine is 0.5.
Since trigonometric functions are periodic, their inverses are not true inverses unless the domain of the trigonometric function is restricted. The principal values for inverse trigonometric functions are defined within specific ranges:
- \(y = {\sin ^{ - 1}}x\), domain: \([-1, 1]\), range: \([-\frac{\pi}{2}, \frac{\pi}{2}]\)
- \(y = {\cos ^{ - 1}}x\), domain: \([-1, 1]\), range: \([0, \pi]\)
- \(y = {\tan ^{ - 1}}x\), domain: \((-\infty, \infty)\), range: \((-\frac{\pi}{2}, \frac{\pi}{2})\)
- \(y = {\csc ^{ - 1}}x\), domain: \((-\infty, -1] \cup [1, \infty)\), range: \([-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]\)
- \(y = {\sec ^{ - 1}}x\), domain: \((-\infty, -1] \cup [1, \infty)\), range: \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\)
- \(y = {\cot ^{ - 1}}x\), domain: \((-\infty, \infty)\), range: \((0, \pi)\)
When solving equations involving inverse trigonometric functions, it's important to be mindful of the principal ranges and the conditions for applying identities like the sum formula for inverse tangents. Failure to check these conditions can lead to including extraneous solutions or excluding valid ones.
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