Question

how to solve tanx +cot x= 8cos2x

Answer 1

\[\tan x+\cot x=8\cos2x~~\text{or}~~\tan x+\cot x=8\cos^2x~~?\]

Answer 2

`tan2x=8cos^2x-cotx`

`sin(2x)/cos(2x)=8cos^2(x)-cos(x)/sin(x)`

Multiply by sin(x)cos(2x)

`sin(2x)sin(x)=8cos^2(x)sin(x)cos(2x)-cos(2x)cos(x)`

Rearanging

`8cos^2(x)sin(x)cos(2x)-cos(2x)cos(x)-sin(2x)sin(x)=0`

Using `sin(2x)=2sin(x)cos(x)`

`8cos^2(x)sin(x)cos(2x)-cos(2x)cos(x)-2sin^2(x)cos(x)=0`

Factor out cos(x)

`cos(x)(8cos(x)sin(x)cos(2x)-cos(2x)-2sin^2(x)) = 0`

Using `2sin^2(x)=1-cos(2x)` and `cos(x)sin(x) = 1/2sin(2x)` from the double angle formulas

`cos(x)(4cos(2x)sin(2x)-cos(2x)-(1-cos(2x))) = 0`

Using `2cos(2x)sin(2x)=sin(4x)`

`cos(x)(2sin(4x)-cos(2x)-1+cos(2x)) = 0`

`cos(x)(2sin(4x)-1) = 0`

so either `cos(x) = 0` or `2sin(4x)-1 = 0`

`cos(x) = 0` at `x=pi/2`

`2sin(4x)-1 = 0`

`sin(4x)=1/2` when `4x=pi/6` or `4x=(5pi)/6`

Gives us the answers `pi/2, pi/24,` and `5pi/24

Answer 3

that's tanx + cotx = 8cos2x

Answer 4

oh sorry my mistake

Answer 5

here let me do it again

Answer 6

can u plz help me solve it?

Answer 7

yes i will

Answer 8

ok thx

Answer 9

timanti?

Answer 10

Left side: sin x/cos x + cos x/sin x = (cos^2 x + sin^2 x)/sin x*cos x = 2/sin 2x.
2/sin 2x = 8cos 2x
2 = 8sin 2x*cos 2x = 4sin 4x
sin 4x = 1/2 = sin Pi/6 and sin 5Pi/6
a) 4x = Pi/6 -> x = Pi/24
b) 4x = 5Pi/6 -> x = 5Pi/24

Answer 11

thank you thu1935