need help with how to solve 2 sin squared x + 3 sin x - 2 = 0?

Answer 1

Maybe you mean

2sin^2(x) + 3sin(x) - 2 = 0
(2sin(x) - 1)(sin(x) + 2) = 0

BBQbaby agrees with my interpretation of the problem. There is no need to use the quadratic formula when the factoring is so easy.

Her analysis of the answers makes sense.

Answer 2

easy peasy, quadratic formula sinX = -b +- square root (b-squared - 4ac) / 2a.

In this case, a = 2, b = 3 and c = -2, therefore
sinX = -3 +- Square root (3-squared - 4 x 2 x(-2)) / 2 x 2
= (-3 +- 5) / 4
= -2 or 0.5
But since a sine ratio must be between 1 and -1, the -2 is meaningless, so the solution is x = 30, as the sine ratio of 30 equals 0.5.

Did I just do your homework for you?

Answer 3

you pose sin x = a discrding all answers > 1 or lower than -1

so 2a^2 +3a -2 =0 roots :-3+((9+16)^0.5))/4 =0.5
The other root is -2 impossible

this gives you sin x = 0.5 and x = 30° or x = 150°

Answer 4

either sin(x+3) = 0 or sin(x-2)=0
(if the product of terms is 0, then at least one of the terms must = 0)

sin(x+3) = 0 if x = n * pi - 3 for each integer n

sin(x-2)=0 if x = n*pi + 2 for each integer n

Answer 5

2sin^2(x+3) sin(x-2) = 0
sin(x + 3) = 0 or sin(x - 2) = 0
x = (n)PI - 3 or (n)PI + 2 where n is an integer.

Answer 6

2s^2+3s-2 = 0

s =(-3+/- sqrt(9+16))/4
s = -2 or (1/2)

if sin x = (1/2)
x is 30 degrees (or 150 degrees)

Answer 7

Look, "trig," either the first two responders were having you on (which you certainly deserved because of your sloppiness), or they simply weren't able to make up for the writing deficiencies in your DREADFULLY posed question.

If you are going to pose what are very probably homework problems, the LEAST you can do is to pose them PROPERLY.

That means: put bloody BRACKETS in your expressions where they are ABSOLUTELY REQUIRED TO AVOID NEEDLESS AMBIGUITY!!