+119 "Simplify using identities so that you can solve the equation. Give the general solution in radians."
sin(x)=1-2cos^2(x)
I ended up getting...
x = - pi/6 + n2pi, x = - 5pi/6 + n2pi, x = pi/2 + n2pi
...which apparently is incorrect. I'd appreciate some help here since I'm not sure how else to solve this.
+129852 sin (x) = 1 - 2cos^2 (x)
Note that cos^2(x) = 1 - sin^2(x)
So we have
sin x = 1 - 2 [ 1 - sin^2 (x) ]
sin x = 1 - 2 + 2sin^2(x)
sin (x) = - 1 + 2sin^2(x) rearrange as
2sin^2(x) - sin (x) - 1 = 0 factor as
(2sin (x) + 1) (sin (x) - 1) = 0
Se each factor to 0 and solve
2sin x + 1 = 0 subtract 1 from both sides
2sin x = -1 divide both sides by 2
sinx = -1/2 and this occurs at 7 pi / 6 + n*2pi and 11pi / 6 + n*2pi
[ where n is an integer]
And..for the other solution we have that
sin x - 1 = 0 add 1 to both sides
sin x = 1 and this occurs at pi/2 + n*2pi
