The fundamental argument

tanx+secx=2cosx

=>sinx/cosx+1/cosx=2(cos (x))^2

=>sinx+1=2(cosx)^2

=>sinx=cos2x

1.Draw the graph of sin x and cos2 x ,the frequency of cos2 x being twice that of sin x,on which we find the graph intersects at two points.Hence two solution.

Substitute with 0,90,180,270,360 (critical points) only 2 will work :D or well algebric way let x is the angle in right angled triangle, whose sides are hyp=z,adj=x,opposite=y tanx= y/z secx= h/z cosx= z/h substitute in equ. y/z + h/z = z/h (y+h)/z = z/h while z^2 + y^2 = h^2 substitute with h (z^2 + y^2 +y ) /z = z/(z^2+y^2) cross cross multiplication z^4 + y^2 + y^3 + 2Z^2y^2 + z^2y -Z^2 = 0 taking common factors z^2(z^2-1) + y^3(y+1) + z^2y(2y+1) = 0 therefore Z^2=0 >> Z=0 Z^2=1, Z= 1,-1, (-1 refused ( length can't be negative ) ) >> z=1 y^3=0 >>> y=0 y=-1 ( refused ) so set solutions is S(z,y)=(0,0),(1,0) therefore number of solutions is 2

(sinX+1)/cosX=2cosX sinX+1=2cos^2X sinX+1=2-2sin^2X 2sin^2X+sinX-1=0 (2sinX-1)(sinX+1)=0 sinX=1/2 or sinX=-1 X=pi/6 or X=3pi/2(ref. because tax=infinity) or X=5pi/6

2 Solutions

2.

tan x + sec x = 2 cos x; sin x + 1 = 2 cos^2 x; 1+sin x = 2 (1 + sin x) (1 - sin x); 1 = 2 - 2 sin x; 2 sin x = 1; sin x = 1/2; x = pi/6 or 5*pi/6; 2 solutions.

sinx=cos2x

tan x+sec x=sin x/cos x+1/cos x=2 cos x, since tan x=sin x/cos x and sec x=cos x =(sin x+1)/cos x=2 cos x sin x+1=2 sqr cos x, sqr cos x=1- sqr sin x sin x+1=2(1- sqr sin x) sin x+2 sqr sin x +1-2=0 sin x+ 2 sqr sin x-1=0, let y=sin x then, y +2 sqr y-1=0, using quadratic formula y=1/2 and y=-1. therefore sine inverse of y= 30 degree and 270 degree.i.e. x=30,x=270 degree

sinx/cosx+1/cosX=2cosx sinx+1=2Cos^2x so we have only 2 possible when used values 0,90,180,270&360