Hello :
<span> sin 2x = sin x and 0 ≤ x ≤ 2π.
all solutions :
2x= x +2k</span>π or x= π -x +2kπ ..... k in : Z
x = 2kπ or : x = π/2 + kπ
but :
<span>0 ≤ x ≤ 2π
</span><span>all values of x such that sin 2x = sin x and 0 ≤ x ≤ 2π are :
</span>k = 0 : x=0 , x= π/2
k=1 : x=2 π , x=3π/2
The answer is B. There can not be any alike "x" inputs.
you will need to use the law of cosines since this picture does not indicate that this is a right triangle
c^2 = a^2 + b^2 – 2ab cos C,
16^2 = 17^2+8^2 -2*17*8*cosC
256=289-272cosC
-33=-272cosC
33/272 = cosC
cos^-1 (33/272)=C taking the inverse cos
C=83.0 (to the nearest tenth)
b^2 = a^2 + c^2 – 2ac cos B,
8^2 = 17^2 +16^2 -2*17*16cosB
64=289-544cosB
-225=-544cosB
225/544=cosB
cos^(-1) =B
B=65.5
A=180-83-65.6
A=31.4
