Question

The number of solutions of the equation $8{\tan }^2\theta +9=6\sec \theta$ in the interval $(\frac{-\pi }{2},\frac{\pi }{2})$

• A. Two
• B. Four
• C. Zero
• D. None of these

Answer 1

The correct option is C Zero

${\left(\tan x\right)}^2$=${\left(\sec x\right)}^2$-1

So, 8 [${\left(\sec x\right)}^2$-1]+9=6$\sec x$

8${\left(\sec x\right)}^2$-8+9=6$\sec x$

6$\sec x$=8${\left(\sec x\right)}^2$+1

8${\left(\sec x\right)}^2$-6$\sec x$+1=0

So, we get a quadratic in $"\sec x".$

on solving we get, there does not exist any value of 'x' for which the aove equation is satisfied.

Therefore, zero solution.

(sec⁡