I am utterly confused with solving these for theta or x. I think we use trig identities to find x?
a. sin^2x = 2sin^2(x / 2)
Note sin^2 (x / 2 ) = 1 - cos x
So we have
sin^2x = 1 - cos x
1 - cos^2x = 1 - cos x subtract 1 from both sides
-cos^2x = -cos x multiply through by -1
cos^2x = cosx
cos^2x - cos x = 0
cosx ( cos x - 1) = 0
Set each factor to 0 and we have that
cos x = 0 and this happens at pi/2 + n*pi where n is an integer
and
cos x - 1 = 0
cos x = 1 and this happens at 0 + n* 2pi where n is an integer
b. sinx - cos^2 x - 1 = 0
sin x - [ 1 - sin^2x ] - 1 = 0
sin^2x + sinx - 2 = 0 factor
(sin x - 1) ( sin x + 2) = 0
Set each factor to 0 and solve for x
sin x - 1 = 0
sin x = 1 and this happens at pi/2 + n* 2pi where n is an integer
And
six + 2 = 0 has no solution
c. 2sin^2 θ + 3sin θ = 2
2sin^2 θ + 3sinθ - 2 = 0 factor
(2sinθ - 1) (sin θ + 2) = 0
Setting each factor to 0....solve for θ
2sinθ - 1 = 0
2sinθ = 1
sinθ = 1/2 and this happens at pi/6 + n*2pi and 5pi/6 + n*2pi where n is an integer
And
sin θ + 2 =0
sinθ = -2 [ this has no solution ]
d. cos (2x) - cos x = 0
2cos^2x - 1 - cos x =0
2cos^2x - cos x - 1 = 0 factor
(2cos x + 1) (cos x - 1) = 0 set each factor to 0 and solve for x
2cos x + 1 = 0
2cos x = -1
cos x = -1/2 and this happens at 2pi/3 + n*2pi and 4pi/3 + n*2pi where n is an integer
And
cos x - 1 = 0
cos x = 1 and this happens at 0 + n*2pi where n is an integer