Question #82a63

Knowing that, `sinx=2sin(x/2)cos(x/2), &, 1+cosx=2cos^2(x/2)`, we have,

`sinx/(1+cosx)=2.`

`rArr {2sin(x/2)cos(x/2)}/(2cos^2(x/2))=2.`

`rArr tan(x/2)=2, .............[if, cos(x/2)ne0,}`

`rArr tan(x/2)=tan(arc tan2)`

`rArr x/2=kpi+arc tan 2, k inn ZZ`.

`rArr x=2kpi+2arc tan2, k in ZZ.`

Considering the constraint `cos(x/2) ne0`, we observe that,

`cos(x/2)=0 rArr 1+cosx=2cos^2(x/2)=0,` & so, the eqn.

becomes meaningless. `:. cos(x/2)!=0.`

`:." The Soln. Set="{2kpi+2arc tan2 : k in ZZ}.`

The Other eqn., `: sinx/(1-cosx)=2,` can similarly be dealt with.

Enjoy maths.!

sin x = 2 + 2cos x
sin x - 2cos x = 2 ( 1 )
Call `tan t = sin t/(cos t) = 2` -->`t = 63^@43`
Equation (1) -->
`sin x - (sin t)/(cos t)cos x = 2cos x = 2(0.45) = 0.90`
sin (x - t) = sin (x - 63.43) = 0.90
Calculator and unit circle give -->2 solutions

a. `(x - 63.43) = 64^@16`
`x = 64.16 + 63.43 = 127^@59`

b. `x - 63.43 = 180 - 64.16 = 115^@84`
`x = 115.84 + 63.43 = 179^@27`

sin x = 2 + 2cos x
sin x - 2cos x = 2
Call `tan t = sin t/(cos t) = 2` --> `t = 63^@43` --> cos t = 0.45
`sin x - (sin t/cos t)cos x = 2`
`sin x.cos t - sin t.cos x = 2cos t`
`sin (x - t) = 2cos t = 0.90` .
Calculator and unit circle give 2 solutions:
a. (x - t) = (x - 63.43) = 64.16 -->
`x = 63.43 + 64.16 = 127@59`
b. (x - 63.43) = 180 - 64.16 = 115.84
`x = 115.84 + 63.43 = 179^@27`
Answers for (0, 360):
`127^@59; 179^@27`
Check by calculator:
x = 127.59 --> sin x = 0.78 --> cos x = - 0.61
sin x - 2cos x = 0.78 + 1.22 = 2. OK
x = 179.27 --> sin x = 0.01 --> cos x = - 0.995 -->
sin x - 2cos x = 0.01 + 1.99 = 2. OK