number of solutions

tanx+secx= 2cosx

multiply by cosx

sinx+ 1 = 2cos²x

sinx+1 = 2- 2sin²x

2sin²x+ sinx - 1 = 0

2sin²x + 2sinx - sinx - 1 = 0

2sinx(sinx+1) - (sinx+1) = 0

(sinx+1)(2sinx-1) = 0

sinx= -1 , sinx = 1/2

so x = 3pie/2 , or x = pie/6 ,or x =pie - pie/6

but for x = 3pie/2 , tanx isnt defined

hence x = pie/6 or 5pie/6

hence 2 solutions

is the ans correct ?

y m i not able to open latex ?

yes sir , i do want to improve my graphing abilities , and m learning the same from Tiit !!! and some frndz wich i made here !!! [1]

ok....i hav to say something [3]

will teh no of point of intersection of grphs sin x and cos (2x) give the no of solns of tan x+ sec x= 2cos x

bec no of point of intersection of grphs sin x and cos (2x) will be 3......3pi/2 will also be its soln

but for this tan x+ sec x= 2cos x 3pi/2 is not a soln.....bec tan(3pi/2) is not defined

OUT OF THESE , one point of intersection is not the solution

u r rong i guess qwerty

sin 270=-1

cos(2*270)=-1

so how can u say 3pi/2 is not a soln of sinx=cos2x

bhai tanx ka kya ??? [5]

Another way:
\tan x + \sec x = 2 \cos x \Rightarrow \sec x - \tan x = \frac{1}{2 \cos x} = \frac{\sec x}{2}

\Rightarrow \sec x = 2 \tan x

From the 1st eqn this implies

\cos x = \pm \frac{\sqrt 3}{2} and cos and tan must both have the same sign.

So 2 solutions