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
.
The correct answer maybe
Actually I don't understand ur question
Answer:
<h3>The solution is 480 bricks.</h3>
Step-by-step explanation:
We are given that number of rows of bricks in a wall = 15 rows.
Number of bricks in a row = 32 bricks.
In order to find the total number of bricks, we need to multiply number of rows with number of bricks in a row.
Total number of bricks in 15 rows = 15 × 32 = 480 bricks.
Therefore, will Chin-li will need 480 bricks.
<h3>The solution is 480 bricks.</h3>
-sqa(3.24-sqa(1.44+sqa(12.25)))
= -sqa(3.24-sqa(4.94)) (since sqa(1.44+sqa(12.25))=sqa(4.94))
=-sqa(3.24-2.22261)
= - sqa(1.01739)
= - 1.00866
