how do i find the exact solution? 2sin^2x+5sinx-3=0 for 0<x<2pie?    there are _ under the< symbol please help

Expert Answers info

embizze eNotes educator | Certified Educator

calendarEducator since 2011

write3,000 answers

starTop subjects are Math, Science, and Business

Solve `2sin^2x+5sinx-3=0` for `0<=x<=2pi` :

`2sin^2x+5sinx-3=0`

`(2sinx-1)(sinx+3)=0`

`sinx+3=0 ==>sinx=-3` but the range of sinx is `-1<=y<=1`

`2sinx-1=0==>sinx=1/2`

The sin is equal to 1/2 at `pi/6 +k2pi,(5pi)/6+k2pi` where k is an integer.

---------------------------------------

For `0<=x<=2pi` we have `x=pi/6,(5pi)/6`

---------------------------------------

The graph:

check Approved by eNotes Editorial