Question

Question #071e3

Answer 1

`x=(7pi)/6+2kpi vv x=(11pi)/6+2kpi,k in ZZ`

Explanation:

If `ab=0`, then either `a=0` or `b=0` or `a=b=0`.

Looking at the equation `(sin(x)+3)(2sin(x)+1)=0`, it can be said that the solutions occur when `sin(x)+3=0` or `2sin(x)+1=0`.

Consider `sin(x)+3=0` first. The range of `sin(x)` is `-1≤sin(x)≤1` for all `x in RR`. There are no solutions here.

Consider `2sin(x)+1=0`. This is equivalent to `sin(x)=-1/2`. Solving for `x` in the range `[0,2pi)`, we get `(7pi)/6` and `(11pi)/6`.

Since `sin(x)` is periodic with period `2pi`, the solution is `x=(7pi)/6+2kpi vv x=(11pi)/6+2kpi,k in ZZ`.

Answer 2

`x=(7pi)/6+2kpi" or "x=(11pi)/6+2kpi`
`color(white)("XXXXXXXXXXXXXXXXXXXXX")AAk in ZZ`

Explanation:

If `(sin(x)+3)(2sin(x)+1)=0`
#{: ("either ",sin(x)+3=0," or ",2sin(x)+1=0), (,"impossible",,rarr sin(x)=-1/2), (,color(white)("X")"since "sin(x) in[-1,+1],,x=(7pi)/6" or "(11pi)/6), (,,,color(white)("X")"assuming "x in [0,2pi)) :}#

Detail:
`color(white)("XXX")sin(x)=-1/2`
implies a reference angle of `pi/6` (one of the standard angles)
in either Quadrant III or IV