2sin^2x - 5sinx = -3?

Question

how would you solve for x?

Answer 1

You can do this in a couple of steps.

First set X = sin x, so that you have the equation 2 X^2 - 5 X = -3. Adding 3 to both sides, you get the equation 2 X^2 - 5 X + 3 = 0, where the polynomial factors as

2 X^2 - 5 X + 3 = (X-1) (2X - 3).

Hence either X = 1 or X= 3/2. Now X = sin x cannot be greater than 1, so clearly sin x = X = 1.

Now for the second step. You want to ask for which x is sin x = 1. This happens only when x = (pi/2) + 2*pi*k for some integer k.

Answer 2

Call sinx = z So 2z^2-5z+3=0 2nd degree equation in z
z= ((5+-sqrt(25-24))/4

z= 3/2 and z= 1 sinx=3/2 can´t be because sinx <=1

z= 1 sinx = 1 x= pi/2 +2kpi k any integer including 0

Answer 3

system

sinsqarex=1
sinx=1

x = pi/2