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
.
Answer:
it's D
Step-by-step explanation:
i just put it into desmos and looked for points, none of the answers appeared, therefore it's D
Answer:
2.4
Step-by-step explanation:
Answer:
i think it is b(-22)
Step-by-step explanation:
i hope this helps
The triangle must be isosceles, due to the perpendicular bisector. Therefore, the expression 20 - 15 can be used to yield an answer of B. 5.
Hope this helps!
