5.05 1. find all solutions to the equation in the interval [0, 2π). cos 2x - cos x = 0 a) 0, two pi divided by three. , four pi divided by three. b) pi divided by six , five pi divided by six , three pi divided by two. c) 0, pi divided by two. , seven pi divided by six. , eleven pi divided by six d) no solution 2. rewrite with only sin x and cos x. sin 3x a) 2 sin x cos2x + cos x b) 2 sin x cos2x + sin3x c) sin x cos2x - sin3x + cos3x d) 2 cos2x sin x + sin x - 2 sin3x 3. find the exact value by using a half-angle identity. cosine of five pi divided by twelve. with answer only if you are sure. you

sin(A+B)=sinAcosB+cosAsinB

sin(x+2x)=sinxcos2x+cosxsin2x

sin2x = 2sinxcosx

cos2x = (cosx)^2 - (sinx)^2 

sin(x+2x)=sinx((cosx)^2 - (sinx)^2)+cosx(2sinxcosx)

we have (sinx)^2 =1- (cosx)^2 

sin(x+2x)=sinx((cosx)^2 - (1- (cosx)^2)+cosx(2sinxcosx)

sin(x+2x)=sinx((2cosx)^2 - 1)+2sinx(cosx)^2

sin(x+2x)=2sinx(cosx)^2 - sinx+2sinx(cosx)^2

(cosx)^2 =1- (sinx)^2 

sin(x+2x)=2sinx(cosx)^2 - sinx+2sinx(1- (sinx)^2)

sin(x+2x)=2sinx(cosx)^2 - sinx+2sinx- 2(sinx)^3)

cos(x/2)=±(√1+cos(x))/2

We know that cos(π/6)=√3/2.

So

ok so the answer to your question is simple as long as you look at the different dimensions that contain the exemplary letters that combined in the right order give you words that eventually within the 5th dimension will conform and give you the correct answer to your question. hope i , stop looking at this easily, look at it with a spectral view.

step-by-step explanation:

told you