1.

Find the number of roots of the equation \( 9 \sec ^{3} \theta-18 \tan ^{2} \theta-\sec \theta-16=0 \) in \( [0,2 \pi] \)

Answer»

\(9sec^3\theta - 18 tan^2\theta - sec\theta - 16 = 0\) 

⇒ \(9sec^3\theta - 18 tan^2\theta+18 - sec\theta - 16 = 0\)  \((\because tan^2\theta = sec^2\theta -1)\)

⇒ \(9sec^3 \theta - 18sec^2\theta - sec\theta + 2 = 0\)

Let \(sec\theta = x\)

Then \(9x^3 - 18x^2 - x + 2 = 0\)

⇒ \((x-2) (9x^2 -1) = 0\)

⇒ \((x -2) (3x -1) (3x + 1) = 0\)

⇒ \(x = 2 \,or\,x = \frac 13 \,or \,x = \frac{-1}3\)

But \(x = sec\theta \in (-\infty, - 1]\cup [1, \infty)\)

So, \(x\ne \frac{-1}3 \) & \( x\ne \frac 13\).

\(\therefore x = 2\)

\(\therefore sec\theta = 2\) is only root of given equation.


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