Question #abd69

Apply these identities, calling `tan (x/2) = t`
`sin x = (2t)/(1 + t^2)`
`cos x = (1 - t^2)/(1 + t^2)`
2sin x - 3cos x = 1
Replace sin x and cos x by the above identities, we get:
`4t - 3 + 3t^2 = 1 + t^2`
`2t^2 + 4t - 4 = 0`
`t^2 + 2t - 2 = 0`.
Solve this quadratic equation for t, using the improved quadratic formula (Google, Yahoo Search);
`D = d^2 = b^2 - 4ac = 4 + 8 = 12` --> `d = +- 2sqrt3`
There are 2 real roots:
`t = -b/(2a) +- d/(2a) = -2/2 +- (2sqrt3)/2 = -1 +- sqrt3`
`t1 = -1 -sqrt3 = - 2.73`
`t2 = -1 + sqrt3 = 0.73`
a. `t1 = tan (x/2) = -2.73` --> `x/2 = - 69^@88` and `x/2 = -69.88 + 180 = 110^@12` --> x = 220.24 + k180
b. t2 = 0.73 --> `x/2 = 36.13` and `x/2 = 36.13 + 180 = 216.13` -->
`x = 72.26 + k180`

Using identities of fractional angle

`sinx=(2tan(x/2))/(1+tan^2(x/2))`

and
`cosx=(1-tan^2(x/2))/(1+tan^2(x/2))`

Given equation becomes

`2sinx-3cosx=1`

`=(4tan(x/2))/(1+tan^2(x/2))-3((1-tan^2(x/2)))/(1+tan^2(x/2))=1`

putting `tan(x/2)=t` we get

`=>(4t)/(1+t^2)-(3(1-t^2))/(1+t^2)=1`

`=>4t-3+3t^2=1+t^2`

`=>3t^2-t^2+4t-3-1=0`

`=>2t^2+4t-4=0`

`=>t^2+2t-2=0`

`=>t^2+2t+1=3`

`=>(t+1)^2=3`

`=>t+1=+-sqrt3`

`=>t=+-sqrt3-1`

when `t =sqrt3-1`

`tan(x/2)=sqrt3-1`

`:.x=2xxtan^-1(sqrt3-1)=2*36.2^@=72.4^@`

Or,`x=(180+72.4)=252.4^@`

Again when

`tan(x/2)=-sqrt3-1`

`=>x=2xxtan^-1(-sqrt3-1)`

`=>x=2*110.1^@=220.2^@`