Question

How do you solve `sin(2x-1)=sin(3x+1)` ?

Answer 1

`x=-2`

Explanation:

If `sin(2x-1)=sin(3x+1)`

Then it follows that

`color(white)(".")2x-1color(white)("d")=color(white)("d")3x+1`

`-1-1color(white)("d")=color(white)("d")3x-2x`

`color(white)("d.dd")-2color(white)("d")=color(white)("d")x`

Answer 2

`x={(2k+1)pi/5 , kinZZ}uu{2kpi-2 , k in ZZ}`

Explanation:

Here,

`sin(2x-1)=sin(3x+1)`

`=>sin(3x+1)-sin(2x-1)=0`

We know that,

`color(red)(sinC-sinD=2cos((C+D)/2)sin((C-D)/2)...to(1)`

Using `(1)` we get

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

`=>cos((5x)/2)sin((x+2)/2)=0`

`=>cos((5x)/2)=0 or sin((x+2)/2)=0`

`(i)cos((5x)/2)=0=>(5x)/2=(2k+1)pi/2,kin ZZ`

`=>5x=(2k+1)pi, kinZZ`

`=>color(blue)(x=(2k+1)pi/5, kin ZZ`

`(ii)sin((x+2)/2)=0=>(x+2)/2=kpi,kinZZ`

`=>x+2=2kpi, kinZZ`

`=>color(blue)(x=2kpi-2,kinZZ`

Hence,

`x={(2k+1)pi/5 , kinZZ}uu{2kpi-2 , k in ZZ}`

Answer 3

`x = - 114^@65 + k360^@`
`x = 36^@ + k360^@`

Explanation:

sin (2x - 1) = sin (3x + 1)
Unit circle and property of sin function give 2 solutions for (2x - 1):
`(2x - 1) = (3x + 1)` and `(2x - 1) = pi - (3x + 1)`
a. (2x - 1) = (3x + 1)
x = -2 radians, or
`x = ((180^@)(- 2))/(3.14) = - 114^@65 + k360^@`
b. `(2x - 1) = pi - (3x + 1) = pi - 3x - 1`
`5x = pi = 180^@`
`x = 180/5 = 36^@ + k360^@`
Check with calculator.
x = - 114.65 --> 2x = - 229.30 --> 3x = - 343.95 -->
1 radians = 180/3.14 = 57^@32 -->
(2x - 1) = - 229.3 - 57.32 = - 286.62 --> sin (-286.62) = 0.96 -->
sin (3x + 1) = sin (- 343.95 + 57.32) = sin (-286.63) = 0.96. Proved.
x = 36 --> sin (2x - 1) = sin (72 - 57.32) = sin (14.68) = 0.25
sin (3x + 1) = sin (108 + 57.32) = sin (165.32) = 0.25. Proved