Question

The principal solution(s) of the trigonometric equation
$\sec \theta =\frac{2}{\sqrt{3}}\text{is/are:}$

• A. $\theta =\frac{\pi }{6}$
• B. $\theta =\frac{7\pi }{6}$
• C. $\theta =\frac{11\pi }{6}$
• D. $\theta =\frac{5\pi }{6}$

Answer 1

Answer: (A), (C)

$\theta =\frac{11\pi }{6}$

We know the principal solutions of a trigonometric equation lies in the interval $[0,2\pi ).$

Now, $\sec \theta >0\forall \theta$ in 1st and 4th quadrant.

In 1st quadrant:
$\sec \theta =\frac{2}{\sqrt{3}}\Rightarrow \theta =\frac{\pi }{6}.$
And, In the 4th quadrant:
$\sec \theta =\frac{2}{\sqrt{3}}\Rightarrow \theta =2\pi -\frac{\pi }{6}=\frac{11\pi }{6}.$

$\text{Hence, principal solution of}\sec \theta =\frac{2}{\sqrt{3}}\Rightarrow \theta =\frac{\pi }{6}\&\frac{11\pi }{6}$