Question

How do you solve `2cos^2x+3sinx=3` in the interval `0<=x<=2pi`?

Answer 1

Solution `x= pi/6,pi/2, (5pi)/6` for `0 <=x<=2pi`

Explanation:

`2cos^2x+3sinx=3 or 2(1-sin^2x)+3sinx-3=0 or 2sin^2x-3sinx+1=0 or (2sinx-1)(sinx-1)=0`. So either `2sinx-1=0 or sinx-1=0` When `2sinx-1=0 :.2sinx=1 :. sinx= 1/2; sin(pi/6)=1/2 and sin(pi-pi/6)=1/2 :. x=pi/6,(5pi)/6` When `sinx-1=0 or sinx=1 ; sin (pi/2)=1 :. x= pi/2`
Solution `x= pi/6,pi/2, (5pi)/6` for `0 <=x<=2pi` [Ans]