Question

Find the critical numbers of -4(cos2tsin2t)

Answer 1

If any individual factor on the left-hand side of the equation is equal to 0, the entire expression will be equal to 0.
(cos2t)=0 (sin2t)=0

Set the first factor equal to 0 and solve.
(cos2t)=0

Take the inverse cosine of both sides of the equation to extract t from inside the cosine.
2t=arccos(0)

Take the inverse cosine of 0 to get (pi)/(2).
2t=(pi)/(2)

Divide each term in the equation by 2.
(2t)/(2)=(pi)/(2)*(1)/(2)

Simplify the left-hand side of the equation by canceling the common factors.
t=(pi)/(2)*(1)/(2)

t=(pi)/(4)
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The cosine function is positive in the 1st and 4th quadrants. To find the second solution, subtract the reference angle from 2pi to find the solution in the 4th quadrant.
2t=2pi-(pi)/(2)

Simplify the expression to find the second solution.
t=(3pi)/(4)

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Set the next factor equal to 0 and solve.
(sin2t)=0
Take the inverse sine of both sides of the equation to extract t from inside the sine.
2t=arcsin(0)

Take the inverse sine of 0 to get 0.
2t=0

Divide each term in the equation by 2.
(2t)/(2)=(0)/(2)

Simplify the left-hand side of the equation by canceling the common factors.
t=(0)/(2)

0 divided by any number or variable is 0.
t=0

The sine function is positive in the 1st and 2nd quadrants. To find the second solution, subtract the reference angle from ` to find the solution in the 2nd quadrant.
2t=pi-0

Simplify the expression to find the second solution.
t=(pi)/(2)

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t=(pi)/(4),(pi)/(2)

Answer 2

how does that rewrite to npi/4?

Answer 3

Use the form acos(bt+c)+d to find the amplitude, period, phase shift, and vertical shift.
y=acos(bt+c)+d ,a=1 ,b=2 ,c=0 ,d=0

The period of the function can be calculated using (2pi)/(|b|), where b is taken from the acos(bt+c)+d form of the function.
\[Period=(2\pi)/(|b|)\]

Replace b with 2 in the formula for period.
\[Period=(2\pi)/(|2|)\]

Solve the equation to find the period.
\[Period=\pi\]

The period of the cos2t function is pi so values will repeat every pi radians in both directions.
\[t=(\pi)/(4)\pi n,(3\pi)/(4)\pi n\]

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For sin2t,
Replace b with 2 in the formula for period.
\[Period=(2\pi)/(|2|)\]

Solve the equation to find the period.
\[Period=\pi\]

The period of the sin2t function is pi so values will repeat every pi radians in both directions.
\[t=~0\pi n~,~(\pi)/(2)\pi n\]

Answer 4

thanks for all your help