Mathematics, 29.12.2019 01:31 svarner2001

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

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Answer from: shortcake8047

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 -coordinate on the unit circle is 0. This happens at , $\pi$ , 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 -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 -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 $0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}$ .

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

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