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Alina [70]
3 years ago
10

Find the solutions to sin2(x) + cos(x) = 1, keeping 0 ≤ x < 2π

Mathematics
1 answer:
frozen [14]3 years ago
3 0
Sin2(x) +cos(x)=1
from the relation: (sin2(x) +cos2(x) =1 )
so , sin2(x)=1-cos2(x)
by subs. in the main eqn.
1-cos2(x) + cos(x) =1
by simplify the eqn.
cos(x) -cos2(x)=0
take cos(x) as a common factor
cos(x)* (1-cos(x))=0
then cos(x)=0 && cos(x)=1
cos(x)=0 if x= pi/2
& cos(x) = 1 if x = 0 , 2*pi
so the solution is x= {0,pi/2 , 2*pi}
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Answer:

First Case:

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Second Case:

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Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

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And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

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2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

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8p^2-12p-24=-2p+1

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Factor:

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Grouping:

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\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

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By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

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