What is the principal solutions of the equation \(\tan x=-\frac{1}{\sqrt{3}}\) ?
Principal Solutions of Trigonometric Equations
Understanding principal solutions in trigonometry is crucial for solving equations. Principal solutions are specific values of the variable (usually an angle) that satisfy a given trigonometric equation within a defined range. For equations involving \(\tan x\), the principal solutions are typically sought in the interval \([0, 2\pi)\).
Equation Analysis for \(\tan x = -\frac{1}{\sqrt{3}}\)
We are given the trigonometric equation: \[ \tan x = -\frac{1}{\sqrt{3}} \] To find the principal solutions, we follow these steps:
- Reference Angle Determination: First, consider the positive value of the tangent, which is \(\frac{1}{\sqrt{3}}\). We know that \(\tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}\). So, the reference angle (let's call it \(\alpha\)) is \(\frac{\pi}{6}\).
- Quadrant Identification: The tangent function \(\tan x\) is negative in two quadrants: the second quadrant and the fourth quadrant. This is because \(\tan x = \frac{\sin x}{\cos x}\), and for \(\tan x\) to be negative, \(\sin x\) and \(\cos x\) must have opposite signs.
- In the second quadrant, \(\sin x > 0\) and \(\cos x < 0\).
- In the fourth quadrant, \(\sin x < 0\) and \(\cos x > 0\).
- Calculation of Solutions in Principal Range:
- Second Quadrant Solution: In the second quadrant, an angle \(x\) is given by \(\pi - \alpha\). \[ x = \pi - \frac{\pi}{6} \] \[ x = \frac{6\pi - \pi}{6} \] \[ x = \frac{5\pi}{6} \]
- Fourth Quadrant Solution: In the fourth quadrant, an angle \(x\) is given by \(2\pi - \alpha\). \[ x = 2\pi - \frac{\pi}{6} \] \[ x = \frac{12\pi - \pi}{6} \] \[ x = \frac{11\pi}{6} \]
- Verification: Both \(\frac{5\pi}{6}\) and \(\frac{11\pi}{6}\) lie within the principal solution range of \([0, 2\pi)\).
- \(\frac{5\pi}{6}\) (which is \(150^\circ\)) is in the second quadrant.
- \(\frac{11\pi}{6}\) (which is \(330^\circ\)) is in the fourth quadrant.
Summary of Principal Solutions
The principal solutions for the equation \(\tan x = -\frac{1}{\sqrt{3}}\) are \(\frac{5\pi}{6}\) and \(\frac{11\pi}{6}\).
| Quadrant | Condition for \(\tan x < 0\) | Formula for Angle \(x\) | Calculated Solution |
|---|---|---|---|
| II (Second) | \(\sin x > 0\), \(\cos x < 0\) | \(x = \pi - \alpha\) | \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\) |
| IV (Fourth) | \(\sin x < 0\), \(\cos x > 0\) | \(x = 2\pi - \alpha\) | \(x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\) |
Therefore, the principal solutions are \(\frac{5\pi}{6}\) and \(\frac{11\pi}{6}\).
Important Questions from Inverse Trigonometric Functions
Solve for x?
sin 2x - 4 sin x + 3 = 0, 0 ≤ x ≤ \(π/2\)
- If \(4 \sin^{-1} x + \cos^{-1} x = \pi\), then what is \(\sin^{-1} x + 4 \cos^{-1} x\) equal to?
- What is \(\cot^2 (\sec^{-1}2) + \tan^2 (\text{cosec}^{-1}3)\) equal to?
- Consider the following in respect of inverse circular functions :
I. \(\sin^{-1}(-x) = -\sin^{-1} x\)
II. \(\cos^{-1}(-x) = \cos^{-1} x\)
III. \(\tan^{-1}(-x) = \pi - \tan^{-1} x\)
IV. \(\cot^{-1}(-x) = \pi - \cot^{-1} x\)
How many of the above are correct ? - What is \(\tan \left[ 2 \tan^{-1} \frac{1}{2} - \frac{\pi}{4} \right]\) equal to ?
