Question

Question #df830

Answer 1

Solution:In interval `0 <= x <= 2pi ; x = pi/2 , x= (7pi)/6 and x = (11pi)/6`

Explanation:

`2sin^2x-sinx-1=0 or 2sin^2x-2sinx +sinx -1=0 or 2sinx(sinx-1) +1(sinx-1)=0 or (2sinx+1)(sinx-1)=0 :.`.Either `2 sinx +1=0 :. sinx = -1/2 ; sin(pi+pi/6) = -1/2 and sin (2pi -pi/6)= -1/2 :. x = (7pi)/6 and x = (11pi)/6`
OR
`sinx-1=0 :.sin x =1 ; sin (pi/2) =1 :. x = pi/2`

Solution: In interval `0 <= x <= 2pi ; x = pi/2 , x= (7pi)/6 and x = (11pi)/6` [Ans]