Let $$2{\sin ^2}x + 3\sin x - 2 > 0$$ and $${x^2} - x - 2 < 0$$ ($$x$$ is measured in radians). Then $$x$$ lies in the interval

Number of solutions of the equation $$\tan x + \sec x = 2\cos x\,$$ lying in the interval $$\left[ {0,2\pi } \right]$$ is:

In this questions there are entries in columns 1 and 2. Each entry in column 1 is related to exactly one entry in column 2. Write the correct letter from column 2 against the entry number in column 1 in your answer book.

$${{\sin \,3\alpha } \over {\cos 2\alpha }}$$ is

Column $${\rm I}$$

(A) positive

(B) negative

Column $${\rm I}$$$${\rm I}$$

(p) $$\left( {{{13\pi } \over {48}},{{14\pi } \over {48}}} \right)$$

(q) $$\left( {{{14\pi } \over {48}},\,{{18\pi } \over {48}}} \right)$$

(r) $$\left( {{{18\pi } \over {48}},\,{{23\pi } \over {48}}} \right)$$

(s) $$\left( {0,\,{\pi \over 2}} \right)$$

Options:-

The equation $$\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0\,$$ In the variable x, has real roots. Then p can take any value in the interval