Solve the equation sin^2x-1/2 sinx-1/2=0 where 0lexle2pi ?

2 Answers
Apr 14, 2018

x=pi/2, (7pi)/6, (11pi)/6

Explanation:

(sinx)^2-1/2sinx-1/2=0

2(sinx)^2-sinx-1=0

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

2sinx+1=0 or sinx-1=0

sinx=-1/2

x=(7pi)/6, (11pi)/6

sinx=1

x=pi/2

Apr 14, 2018

x = pi/2

Explanation:

1/ sin^2x - 1/2 sinx - 1/2
2/ Factor it = (sinx + 1/2)(sinx-1)
3/ sinx = -1/2 ; sinx = 1
4/ x = sin^-1(1/2) ; x = sin^-1 (1)
5/ x = pi/6 ; x = pi/2
6/ Check both answer with your calculator and see which one works