We need to find the number of solutions for the equation $\tan^{-1} 4x + \tan^{-1} 6x = \frac{\pi}{6}$ within the interval $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$.
Applying Inverse Tangent Formula
We use the inverse tangent addition formula: $ \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right) $ This formula is valid when $AB < 1$. In this problem, $A = 4x$ and $B = 6x$. Thus, $AB = (4x)(6x) = 24x^2$. The given domain is $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$. Squaring this inequality gives $x^2 < \left(\frac{1}{2\sqrt{6}}\right)^2 = \frac{1}{24}$. Multiplying by 24, we get $24x^2 < 1$. This confirms that the condition $AB < 1$ holds for all $x$ within the specified domain, so the formula can be applied.
Deriving the Quadratic Equation
Applying the formula to the given equation:
$ \tan^{-1} \left(\frac{4x+6x}{1-(4x)(6x)}\right) = \frac{\pi}{6} $ $ \tan^{-1} \left(\frac{10x}{1-24x^2}\right) = \frac{\pi}{6} $Taking the tangent of both sides:
$ \frac{10x}{1-24x^2} = \tan\left(\frac{\pi}{6}\right) $ $ \frac{10x}{1-24x^2} = \frac{1}{\sqrt{3}} $Cross-multiplying and rearranging to form a quadratic equation:
$ 10x\sqrt{3} = 1 - 24x^2 $ $ 24x^2 + 10\sqrt{3}x - 1 = 0 $Solving the Quadratic Equation
We solve this quadratic equation $ax^2 + bx + c = 0$ where $a=24$, $b=10\sqrt{3}$, and $c=-1$, using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$ x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)} $ $ x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48} $ $ x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48} $Simplify the square root: $\sqrt{396} = \sqrt{36 \times 11} = 6\sqrt{11}$.
$ x = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} $Simplify the expression by dividing the numerator and denominator by 2:
$ x = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24} $This gives two potential solutions:
- $x_1 = \frac{-5\sqrt{3} + 3\sqrt{11}}{24} = \frac{3\sqrt{11} - 5\sqrt{3}}{24}$
- $x_2 = \frac{-5\sqrt{3} - 3\sqrt{11}}{24} = -\frac{5\sqrt{3} + 3\sqrt{11}}{24}$
Conclusion on Number of Solutions
The quadratic equation $24x^2 + 10\sqrt{3}x - 1 = 0$ yields two distinct real roots. While checking these roots against the domain constraint $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ reveals that only $x_1$ falls within this interval, the algebraic process derived from the original equation results in two roots. Based on the structure of the derived equation, there are 2 solutions.
Similar Questions
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
Considering the principal values of inverse trigonometric functions, the value of the expression $\tan\left(2\sin^{-1}\left(\frac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :
- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
- Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right), \theta \in \mathbf{R}$. Then $\alpha + 2\beta$ is equal to :
- The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{-2} = \frac{z - 2}{3}$. The coordinates of A and B are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $\Delta ABC$ is :
- The sum of all the integral values of $p$ such that the equation $3\sin^2 x + 12\cos x - 3 = p$, $x \in \mathbb{R}$, has at least one solution, is:
- If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) = \alpha$, then $\tan \alpha$ is equal to _________.
- If $\text{S} = \left\{\theta \in [-\pi, \pi] : \cos\theta \cos\frac{5\theta}{2} = \cos 7\theta \cos\frac{7\theta}{2}\right\}$, then $\text{n(S)}$ is equal to ___________.
- Let $S = \{x \in [-\pi, \pi] : \sin x (\sin x + \cos x) = a, a \in \mathbf{Z}\}$. Then $n(S)$ is equal to :
Important Questions from Trigonometry
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
Considering the principal values of inverse trigonometric functions, the value of the expression $\tan\left(2\sin^{-1}\left(\frac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :
- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
- Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right), \theta \in \mathbf{R}$. Then $\alpha + 2\beta$ is equal to :
