Question

find all values of x in the interval [0,2pi] that satisfy,
a) sin^2x+sin x-2=0
b) cosx(sinx-(1/2))=0
Please explain your steps to me so I can understand how to so this, my trig identities are not so great. Thank you:)

Answer 1

let sin(x) = y,this is the equation right \[y ^{2}+y-2 = 0\]
solutions is y = 1 or -2 right
but y is sin(x), sin(x) can be in -1 to 1 only, so rule out -2

Answer 2

now find all the possible values for sin(x) =1 in 0 to pi

Answer 3

2nd one, either cos(x) =0 or sin(x) = 1/2,
cos(x) = 0 would imply x= pi/2 or 3pi/2
sin(x) = 1/2 would imply x= pi/6 or 5pi/6 or 7pi/6 or 11pi/6

Answer 4

theloser covered the structure of the solution pretty well, but make sure to read the problem. It wanted the solutions in the domain [0, 2pi]. I think sin x is zero at x=0, x=pi, or x=2pi in that interval.

Answer 5

For the second problem, theloser's analysis is good, but the third and fourth answers on the last line don't work (at that point, sin x is -1/2).

Answer 6

Thank you for your help. I appreciate it.