Question

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.

PLEASE HELP WITH ANSWER ONLY IF YOU ARE SURE. THANK YOU

Answer 1

1. Find all solutions to the equation in the interval [0, 2π).
cos 2x - cos x = 0
using a graphical tool
x1=0
x2=2π/3
x3=4π/3
the answer is the letter A) 0, two pi divided by three. , four pi divided by three.

2. Rewrite with only sin x and cos x.
sin 3x
sin(3x)=sin(2x+x)
sin(A+B)=sinAcosB+cosAsinBsin(x+2x)=sinxcos2x+cosxsin2xsin2x = 2sinxcosxcos2x = (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)
sin(x+2x)=2sinx(cosx)^2 +sinx- 2(sinx)^3)
the answer is the letter 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

cos(x/2)=±(√1+cos(x))/2 We know that cos(π/6)=√3/2. So cos(π/12)=(√2+√3)/2 See attached file problem 13
the answer is one half times the square root of quantity two plus square root of three