Question

Need someone to check this for me?
@vocaloid
@smokeybrown
@angle

Answer 1

for 10 the sol'n is right but I would include a few more steps b/c it's not clear how you got x = 2pi*n from 2-2cos(x)-sin^2(x) = 0

you could substitute sin^2(x) for 1 - cos^2(x) to get

2 - 2cos(x) - (1 - cos^2(x)) = 0

cos^2(x) - 2cos(x) + 1 = 0
then factor that, and show that x = 2pi*n using the unit circle

Answer 2

11 is also correct but there's a key step missing, you need to expand 2sin(2x) to get 2[2cos(x)sin(x)]

and ~then~ it will factor into (2cos(x)-1)(2sin(x)+sqrt(3)) = 0

Answer 3

12, also good, they might want you to write out that

csc(4x) = 1/sin(4x) = 2 giving us sin(4x) = 1/2 and x = pi/24 and 5pi/24 using the UC

Answer 4

13. cos^2(2x) - sin^2(2x) = cos(4x) = 0

cos equals 0 when the angle = pi/2 or 3pi/2 so

dividing this by 4 we get

pi/8 + 2pi*n and 3pi/8 + 2*pi*n because cos repeats itself every 2pi*n units not every pi*n

Answer 5

14 is almost good, the last two solutions are good

since sin(2x) = sin(x) when sin(x) = 0 the other solutions are 0 + 2pi*n and pi + 2pi*n

Answer 6

lmk if anything was unclear

Answer 7

that was very clear

Answer 8

hm.

Answer 9

multiplying sin^2(x) * cos^2(x) gives us [1 - cos^2(2x) ] / 4, still figuring out where to go from there

Answer 10

tup that is what i got so far, stuck on the rest

Answer 11

ok, I got it, you re-use the identity cos^2(u) = [1 + cos(2u)]/2 to re-write

[1 - cos^2(2x) ] / 4

as

[1 - (1 + cos(4x)/2] / 4

multiplying the numerator and denominator by 2 gives us

[2 - (1 + cos(4x)]/8

= [1 - cos(4x)]/8 thus proving the original statement