Question

$\mathrm{The}\mathrm{principle}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{trigonometric}\mathrm{equation}\sec \theta =\frac{2}{\sqrt{3}}\mathrm{are}$

• A.
  $\frac{\pi }{6}$
• B.
  $-\frac{\pi }{6}$
• C.
  $\frac{11\pi }{6}$
• D.
  $\frac{13\pi }{6}$

Answer 1

Answer: (A), (C)

$\frac{11\pi }{6}$
We know the principle solution of a trigonometric equation lies in the interval $[0,2\pi )$.

For, $\sec \theta =\frac{2}{\sqrt{3}}$,
we know $\sec \theta$ is positive only in 1st and 4th quadrant in the interval $[0,2\pi )$.
Now in 1st quadrant
$\sec \theta =\frac{2}{\sqrt{3}}\Rightarrow \theta =\frac{\pi }{6}.$
And, in 4th quadrant
$\sec \theta =\frac{2}{\sqrt{3}}\Rightarrow \theta =2\pi -\frac{\pi }{6}=\frac{11\pi }{6}.$
$\mathrm{Thus},\mathrm{the}\mathrm{principle}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{equation}\sec \theta =\frac{2}{\sqrt{3}}\mathrm{is}\theta =\frac{\pi }{6}\mathrm{and}\frac{11\pi }{6}.$