solve 2x + 5 - 2secxtanx = 0?

Question

find x

how do i do this

thanks

Answer 1

2x+5-2 sinx/cos^2x =0
1)take y= 2x+5 -2sinx/cos^2x which is discontinuos at x=(2k+1)pi/2
Let´s limit to the interval 0,2pi
At x=pi/2 the limit is -infinity and at x=3pi/2 the limit is +infinity
y´=2-2* 1/cos^4(x)* [cos^3( x)+2sin^2x cosx]
y´= 2[1-1/cos ^3 x *(cos^2x +2sin^2x)]=0

1-1/cos^3x (1+sin^2x) =0 so cos^3 x =2-cos^2x

cos^3x +cos^2x-2=0 Which factored =(cosx-1)*(cos^2+2cos+2)= 0 Only root cos x=1 and x=0 0r 2pi

y´= 1/cos^3 x*( cos^3 x+cos^2x-2)
and the sign depends on cos x and (cosx-1 )<=0

0------------pi/2+++++++++3pi/2 -----------2pi
Atx=0 y(0)=5 >0 and till pi/2 is decreasing so there is one root as lim y x=>pi/2 =-infinity
There is also one root in [pi/2,3pi/2] (increasing from -infinity to + infinity
At x=2pi y(2pi)=4pi+5 >0 so there is no root in this interval.
This repeats in the nexts intervals (2pi,4pi) and so on