What are the All the Unit circle solutions to ---> (2sin^2x) + (3sinx) - 2 = 0?

Answer 1

First, convert

[sin(x)]^2 + 3sin(x) - 2 = 0 (I just rewrote your question since this is based on what the question means based on limited text we have to work with.

We solve this like we would any quadratic, since if we substitute y = sin(x), we get

2y^2 + 3y - 2 = 0
(2y - 1 ) (y + 2 ) = 0

2y - 1 = 0
y + 2 = 0

Substituting y = sin(x) back, we get
2sin(x) - 1 = 0
sin(x) = -2

sin(x) = 1/2
sin(x) = -2

We can discard the second equation since -1 <= sin(x) <= 1 [i.e. sin(x) has to lie between -1 and 1], so we work with the first one.

We then ask ourselves: On the unit circle, where is sin(x) equal to 1/2? It is equal to 1/2 at pi/6 and 5pi/6. So

x = pi/6 and 5pi/6

Answer 2

{thank you for employing parentheses to make sparkling the expression.} Multiply the two factors via [(x-2)×(x+3)]: (x + 3) + (x - 2) > 0 ... yet it rather is genuine on situation that (x - 2)(x + 3) is effectual. it rather is, x < -3 or x > 2. 2x + a million > 0 2x > -a million x > -½ ... If x < -3, then x > -½, it incredibly is senseless, subsequently x won't be able to be < -3. even nevertheless, if x > 2 then it particularly is > -½ What happens if -3 < x < 2 ? Then 2x + a million < 0 2x < -a million x < -½ putting those on an identical time -3 < x < -½, or x > 2.