solve for t:
0=4sin(2t)cos(2t)-sin(2t)
0 = sin(2t)[4 cos(2t) -1]
There are two possible solutions:
sin(2t) = 0
(which leads to t = 0 or pi/2)
and
cos(2t) = 1/4
which leads to other values of t.
t = (1/2)cos^-1(1/4)
if cos2t=1/4 then t= what??
Cos inverse of 1/4 =2t
So do cos inverse of 1/4=75.52...
So...
75.52...= 2t
Solve for t.
T=37.76..
To solve this equation for t, we can first factor out sin(2t) from both terms on the right side:
0 = sin(2t)(4cos(2t) - 1)
Now we have a product of two factors equal to zero. According to the zero product property, for the equation ab = 0, either a = 0 or b = 0.
Applying this to our equation, we have two possibilities:
1) sin(2t) = 0
2) 4cos(2t) - 1 = 0
Let's solve each possibility separately:
1) sin(2t) = 0
To solve this, we need to find the values of t that make sin(2t) equal to zero. Recall that sin(x) = 0 when x is an integer multiple of π. Therefore:
2t = nπ (where n is an integer)
t = nπ/2
So, one possible solution is t = nπ/2, where n is an integer.
2) 4cos(2t) - 1 = 0
To solve this equation, we need to isolate cos(2t):
4cos(2t) = 1
cos(2t) = 1/4
Now, we could use the inverse cosine function (also known as arccos or cos^(-1)) to solve for 2t:
2t = arccos(1/4)
However, keep in mind that the inverse cosine function gives us a value between 0 and π (or 0 and 180°) only. To find all possible solutions for 2t, we need to consider the periodicity of the cosine function.
The cosine function repeats its values every 2π (or 360°). Therefore, to find all solutions for 2t, we can use the general solution:
2t = arccos(1/4) + 2nπ
or
2t = -arccos(1/4) + 2nπ
where n is an integer.
Finally, we can divide both sides by 2 to solve for t:
t = (arccos(1/4) + 2nπ)/2
or
t = (-arccos(1/4) + 2nπ)/2
These are the solutions for t that satisfy the given equation.