Question

PLEASE HURRY I NEED HELP SOS!!!!!!!! (10 POINTS)

Answer 1

Answer:

$\theta \in \{0, \pi, \frac{11\pi}{6}\}$

Step-by-step explanation:

$\sin\theta + 1 = \cos(2\theta)\\\sin\theta + 1 = \cos^2\theta - \sin^2\theta\\\sin\theta + \sin^2\theta + \cos^2\theta = cos^2\theta -\sin\theta^2\\2\sin^2\theta +\sin\theta = 0\\\sin\theta[2\sin\theta + 1] = 0\\\sin\theta = 0 || 2\sin\theta + 1 = 0\\\theta \in \{0, \pi, \frac{11\pi}{6}\}$

Answer 2

Answer:

$\{0,\pi , \frac{7\pi}{6},\frac{11\pi}{6}\}$

Step-by-step explanation:

We are going to have to use a double angle identity.

$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$

I'm going to write this in terms of sine since the left hand side of the equation is also in terms of sine.

So applying the Pythagorean Identity $\sin^2(\theta)+\cos^2(\theta)=1$ to the identity above gives you:

$\cos(2\theta)=(1-\sin^2(\theta))-\sin^2(\theta)$

$\cos(2\theta)=1-2\sin^2(\theta)$

So the equation you have becomes:

$\sin(\theta)+1=1-2\sin^2(\theta)$

Get one side to be 0.

I'm going to move everything on right side to left side.

When you move a term over from one side to another, you just need to change the sign in front it. so if is plus 1 on the right it is minus 1 on the left.

If it is minus $2\sin^2(\theta)$ then it is plus $2\sin^2(\theta)$ on the other side.

$2\sin^2(\theta)+\sin(\theta)+1-1=0$

Combine the like terms (1-1):

$2\sin^2(\theta)+\sin(\theta)+0=0$

$2\sin^2(\theta)+\sin(\theta)=0$

Factor out the common factor on the left which is sin( )[/tex]

$\sin(\theta)[2\sin(\theta)+1]=0$

If you have a product is 0 then one of the factors must be zero (both could be 0 also).

$\sin(\theta)=0$ or $2\sin(\theta)+1=0$

I'm going to solve the first equation first.

$\sin(\theta)=0$ when the y-coordinate on the unit circle is 0.

This happens at $0 \text{ or } \pi$ is the given interval you provided.

Let's solve the other equation:

$2\sin(\theta)+1=0$

Subtract 1 on both sides:

$2\sin(\theta)=-1$

Divide both sides by 2:

$\sin(\theta)=\frac{-1}{2}$

So you are looking for when the y-coordinate is -1/2 on the unit circle.

This happens at $\frac{7\pi}{6} \text{ or } \frac{11\pi}{6}$

So the solution set is:

$\{0,\pi , \frac{7\pi}{6},\frac{11\pi}{6}\}$