by the double angle identity for sine. Move everything to one side and factor out the cosine term.
Now the zero product property tells us that there are two cases where this is true,
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of
, so
where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval
for
and
. More generally, if you think of
as a point on the unit circle, this occurs whenever
also completes a full revolution about the origin. This means for any integer
, the general solution in this case would be
and
.
<span>x=<span>−<span><span>1<span> or </span></span>x</span></span></span>=<span>−<span>3 i think lol</span></span>
answer
10
step-by-step explanation
the equation given is sin(x) = cos(y) with x = 2k + 3 and y = 6k + 7
substitute in 2k+ 3 for x in sin(x) and substitute in 6k + 7 for y in cos(y)
sin(x) = cos(y)
sin(2k + 3) = cos(6k + 7)
we know that sin(x) = cos(90 -x)
sin(2k + 3)
= cos(90 - (2k + 3) )
= cos(90 - 2k - 3)
= cos(87 - 2k)
substitute this into the equation sin(2k + 3) = cos(6k + 7)
sin(2k + 3) = cos(6k + 7)
cos(87 - 2k) = cos(6k + 7)
87 - 2k = 6k +7
80 = 8k
k = 10
Answer:
a6b40c70
Step-by-step explanation:
a6b40c70
