The principal solutions of cot x = sqrt{3} ar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation is $\cot x = \sqrt{3}$,which is equivalent to $	an x = \frac{1}{\sqrt{3}}$.We know that $	an \frac{\pi}{6} = \frac{1}{\sqrt{3

Vedclass · Accepted Answer

$\frac{\pi}{6}, \frac{7 \pi}{6}$

Vedclass · Answer

(B) The given equation is $\cot x = \sqrt{3}$,which is equivalent to $	an x = \frac{1}{\sqrt{3}}$.We know that $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.Since $	an x$ is positive in the first and third quadrants,the principal solutions lie in the interval $[0, 2\pi)$.The first solution is $x = \frac{\pi}{6}$.The second solution in the third quadrant is $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.Thus,the principal solutions are $\frac{\pi}{6}$ and $\frac{7\pi}{6}$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

The principal solutions of $\cot x = \sqrt{3}$ are

Similar Questions

The probability distribution of a random variable $X$ is given below:
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $\frac{2}{10}$ $\frac{3}{10}$ $\frac{4}{10}$

Then the variance of $X$ is

Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18} \, J$. Calculate the ionization enthalpy of atomic hydrogen in terms of $J \, mol^{-1}$.

Solid targets of different elements are bombarded by highly energetic electron beams. The frequency $(f)$ of the characteristic $X-$rays emitted from different targets varies with atomic number $(Z)$ as

Three point charges are placed at the corners of an equilateral triangle. Assuming only electrostatic forces are acting,which of the following statements is true?

Which of the following oxides does not form an acidic aqueous solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module