Question

How do you solve `5sin(2x)=1`?

Answer 1

`x=kpi+alpha,kinZandalpha=tan^(-1)(5+-2sqrt6)`
OR
`x=(kpi)/2+(-1)^k*beta/2,kinZandbeta=sin^(-1)(1/5)`

Explanation:

`5sin2x=1=>sin2x=1/5=>(2tanx)/(1+tan^2x)=1/5`
`=>10tanx=1+tan^2x=>tan^2x-10tanx+1=0`
`=>tan^2x-10tanx+25-24=0`
`=>(tanx-5)^2=24=(2sqrt(6))^2`
`=>tanx-5=+-2sqrt6=>tanx=5+-2sqrt6inR`
`=>tanx=tan(tan^(-1)(5+-2sqrt6))`, let ,`alpha=tan^(-1)(5+-2sqrt6)`
`:.x=kpi+alpha,kinZandalpha=tan^(-1)(5+-2sqrt6)`
OR
`5sin2x=1=>sin2x=1/5in[-1,1]`
`=>sin2x=sin(sin^(-1)(1/5))`, let ,`beta=sin^(-1)(1/5)`
`:.2x=kpi+(-1)^k*beta,kinZand beta=sin^(-1)(1/5)`
`=>x=(kpi)/2+(-1)^k*beta/2,kinZandbeta=sin^(-1)(1/5)`

Answer 2

`x = 5^@77 + k180^@`
`x = 84^@23 + k180^@`

Explanation:

`sin 2x = 1/5`
Calculator and unit circle give 2 solutions:
a. `2x = 11^@54 + k360^@` -->
`x = 5^@77 + k180^@`
b. `2x = 180 - 11.54 = 168^@46 + k360^@` -->
`x = 84.23 + k180^@`