The number of solutions of the equation `tanx+secx=2cosx` lying in the interval `[0,2pi]` is 0 (b) 1 (c) 2 (d) 3

Options

A. 0
B. 1
C. 2
D. 4

Answer

c

Explanation

`tan x + sec x = 2 k cos x, x notin (2n + 1 ) (pi)/(2)`
` rArr " " sinx + 1 = 2 cos ^(2) x `
` rArr " " sinx + 1 = 2 ( 1- sin^(2) x )`
` rArr 2 sin ^(2) x + sin x - 1 = 0`
` rArr " " sinx = (1)/(2), sinx = - 1 `
`rArr" "x = (pi)/(6), ( 5pi)/(6)`
or `" " x = ( 3pi)/(2)`
but `" " x notin ( 2 n + 1 ) (pi)/(2)`
` therefore " " x = (pi)/(6), ( 5pi)/(6)`
Here, number of solutions are two.