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iren [92.7K]
4 years ago
7

Find the real zeros of the trigonometric function on the interval 0 ≤ x < 2π

Mathematics
2 answers:
Luda [366]4 years ago
6 0
4cos2(x)−3=0

or 
4cos2(x)=3

cos(x)= √3/2

π/6, 5π/6,  7π/6,  11π/6



Romashka-Z-Leto [24]4 years ago
5 0
F(x) = 4 [cos (x)]^2 - 3 = 0

4[cos(x)]^2 = 3

cos(x) = √3 / 2

That happens in the first and fourth quadrants, for the angles 30 degrees and 330 degrees.

Answer: x = 30 degrees and x = 330 degrees
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3 years ago
Differential Equations Problem
fgiga [73]

Answer:

y=\frac{t^2e^{2t}}{3}+ce^{2t}

Step-by-step explanation:

We have given differential equation \frac{dy}{dt}-2y=t^2e^{2t}

We know that linear differential equation is given by \frac{dy}{dt}+Py=Q

On comparing with standard equation P = -2 and Q= t^2e^{2t}

Now integrating factor IF=e^{-Pdt}

IF=e^{-2dt}=e^{-2t}

Now solution of differential equation is given by

y\times IF=\int\ IF\times Q\ dt

y\times e^{-2t}=\int\ e^{-2t}\times t^2e^{2t}\ dt

y\times e^{-2t}=\frac{t^2}{3}+c

y=\frac{t^2e^{2t}}{3}+ce^{2t}

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39%

Step-by-step explanation:

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