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Aloiza [94]
3 years ago
6

Solve sinX+1=cos2x for interval of more or equal to 0 and less than 2pi

Mathematics
2 answers:
Igoryamba3 years ago
8 0

Answer:

Question 1: \sin(x)+1=\cos^2(x)

Answer to Question 1: x=0, \pi \frac{3\pi}{2}

Question 2: \sin(x)+1=\cos(2x)

Answer to Question 2: 0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}

Question:

I will answer the following two questions.

Condition: 0\le x

Question 1: \sin(x)+1=\cos^2(x)

Question 2: \sin(x)+1=\cos(2x)

Step-by-step explanation:

Question 1: \sin(x)+1=\cos^2(x)

Question 2: \sin(x)+1=\cos(2x)

Question 1:

\sin(x)+1=\cos^2(x)

I will use a Pythagorean Identity so that the equation is in terms of just one trig function, \sin(x).

Recall \sin^2(x)+\cos^2(x)=1.

This implies that \cos^2(x)=1-\sin^2(x). To get this equation from the one above I just subtracted \sin^2(x) on both sides.

So the equation we are starting with is:

\sin(x)+1=\cos^2(x)

I'm going to rewrite this with the Pythagorean Identity I just mentioned above:

\sin(x)+1=1-\sin^2(x)

This looks like a quadratic equation in terms of the variable: \sin(x).

I'm going to get everything to one side so one side is 0.

Subtracting 1 on both sides gives:

\sin(x)+1-1=1-\sin^2(x)-1

\sin(x)+0=1-1-\sin^2(x)

\sin(x)=0-\sin^2(x)

\sin(x)=-\sin^2(x)

Add \sin^2(x) on both sides:

\sin(x)+\sin^2(x)=-\sin^2(x)+\sin^2(x)

\sin(x)+\sin^2(x)=0

Now the left hand side contains terms that have a common factor of \sin(x) so I'm going to factor that out giving me:

\sin(x)[1+\sin(x)]=0

Now this equations implies the following:

\sin(x)=0 or 1+\sin(x)=0

\sin(x)=0 when the y-coordinate on the unit circle is 0. This happens at 0, \pi, or also at 2\pi. We do not want to include 2\pi because of the given restriction 0\le x.

We must also solve 1+\sin(x)=0.

Subtract 1 on both sides:

\sin(x)=-1

We are looking for when the y-coordinate is -1.

This happens at \frac{3\pi}{2} on the unit circle.

So the solutions to question 1 are 0,\pi,\frac{3\pi}{2}.

Question 2:

\sin(x)+1=\cos(2x)

So the objective at the beginning is pretty much the same. We want the same trig function.

\cos(2x)=\cos^2(x)-\sin^2(x) by double able identity for cosine.

\cos(2x)=(1-\sin^2(x))-\sin^2(x) by Pythagorean Identity.

\cos(2x)=1-2\sin^2(x) (simplifying the previous equation).

So let's again write in terms of the variable \sin(x).

\sin(x)+1=\cos(2x)

\sin(x)+1=1-2\sin^2(x)

Subtract 1 on both sides:

\sin(x)+1-1=1-2\sin^2(x)-1

\sin(x)+0=1-1-2\sin^2(x)

\sin(x)=0-2\sin^2(x)

\sin(x)=-2\sin^2(x)

Add 2\sin^2(x) on both sides:

\sin(x)+2\sin^2(x)=-2\sin^2(x)+2\sin^2(x)

\sin(x)+2\sin^2(x)=0

Now on the left hand side there are two terms with a common factor of \sin(x) so let's factor that out:

\sin(x)[1+2\sin(x)]=0

This implies \sin(x)=0 or 1+2\sin(x)=0.

The first equation was already solved in question 1. It was just at x=0.

Let's look at the other equation: 1+2\sin(x)=0.

Subtract 1 on both sides:

2\sin(x)=-1

Divide both sides by 2:

\sin(x)=\frac{-1}{2}

We are looking for when the y-coordinate on the unit circle is \frac{-1}{2}.

This happens at \frac{7\pi}{6} or also at \frac{11\pi}{6}.

So the solutions for this question 2 is 0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}.

stepan [7]3 years ago
7 0

Answer:

thats the first page and the second page

the answer is x {-π/2+1+2kπ}

x={-1/3+π/6+2kπ}

i hope it helps:)

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