Question

How to solve `4sin^2((3x+pi)/6)=3`?

So, `sin((3x+pi)/6)=+-sqrt3/2`. Then what?

Answer 1

`x=(2k+1)pi`v`(2k-1/3)pi`

Explanation:

We have `sin(2pi/3)=sqrt3/2` (see https://socratic.org/questions/what-is-sin-2pi-3-equal-to)

Therefore
`sin((3x+pi)/6)=+-sin(2pi/3)`
`(3x+pi)/6=(2/3pi +2kpi)` v `(-2/3pi +2kpi)`
`-2/3pi` should have the same sin value as `4/3pi`
so that we get
`3x+pi=6(2/3pi +2kpi)` v `6(4/3pi +2kpi)`
`3x=(4pi-pi+12kpi)`v`(12pi-pi+12kpi)`
`3x=(3pi+12kpi)`v`(11pi+12kpi)`
`x=(pi+4kpi)` v `(11/3pi+4kpi)`
`x=(2k+1)pi` v `(2k-1/3)pi`

Answer 2

`x = (2k + 1)pi/3`

Explanation:

`4sin^2 ((3x + pi)/6) = 3`
`sin ((3x + pi)/6) = +- sqrt3/2`
a. `sin ((3x + pi)/6) = sqrt3/2`
Trig table and unit circle give 2 solutions for `(3x + pi)/6`:
1. `(3x + pi)/6 = pi/3` --> `3x + pi = 2pi`
`3x = 2pi - pi = pi + 2kpi` -->`x = pi/3 + (2kpi)/3`
2. `(3x + pi)/6 = (2pi)/3` --> `3x + pi = 4pi`
`3x = 3pi + 2kpi` --> `x = pi + (2kpi)/3`
General answer: `x = (2k + 1)pi/3`
b. `sin ((3x + pi)/6) = - sqrt3/2`
Trig table and unit circle give 2 solutions for `(3x + pi)/6`
1. `(3x + pi)/ 6 = (4pi/3)` --> `(3x + pi) = 8pi`
`3x = 7pi + 2kpi` --> `x = (7pi)/3 + (2kpi)/3`, or
`x = pi/3 + (2kpi)/3`
2. `(3x + pi)/6 = (5pi)/3` --> `3x + pi = 10pi`
`3x = 9pi + 2kpi` --> `x = 3pi + (2kpi)/3`, or
x = pi + (2kpi)/3
General answer: `x = (2k + 1)pi/3`