tan(2x)+2cosx=0?

Question

solve for x in radians, between 0 and 2pi

Answer 1

pi/4 and -pi/6

Answer 2

x <> pi/4 + k*pi/2
2sinx*cosx/(1-2sin^2 x) + 2 cosx = 0
<=> 2cosx [sinx/(1-2sin^2 x) +1] = 0
<=> cosx = 0
or sinx + 1-2sin^2 x =0
<=> x= pi/2 + m*pi ( not suitable with the condition above )
or 2sin^2 x -sinx -1 =0 (1)

(1) <=> sinx =1 or sinx = -1/2
<=> x = pi/2 + n * 2pi (not suitable) or x = -pi/3 + n*2pi
or x= 4pi/3 + n*2pi

So the roots are x = -pi/3 and x = 4pi/3

Answer 3

tan(2x)+2cosx=0

2sinxcosx/(1-2sinxsinx)+ 2cosx=0

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

sinx/(1-2sinxsinx)= -1

2sinxsinx -sinx -1=o

=> sinx = 1 or -1/2

Answer 4

sin(2x)/cos(2x)+2cosx=0
sin2x+2cosxcos2x=0
2sinxcosx+2cosxcos2x=0
2cosx(sinx+cos2x)=0
2cosx=0
cosx=0
x=(2n+1)pi/2

sinx+cos2x=0
sinx+1-2sin^2x=0
2sin^2x-sinx-1=0
2sin^2x-2sinx+sinx-1=0
2sinx(sinx-1)+1(sinx-1)=0
(sinx-1)(2sinx+1)=0
sinx=1
sinx=sinpi/2
x=npi+(-1)^n*pi/2

2sinx+1=0
2sinx=-1
sinx=-1/2
simx=sin(-pi/6)
x=npi+(-1)^n(-pi/6)