Mathematics, 18.03.2021 20:10 kennyg02

I’m extremely stressed and anxious again so please help me pre cal i hate how i don’t understand

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Answer from: mangociruelo9402

$x=0,\\x=\frac{2\pi}{3},\\x=\pi\\x=2\pi$

Step-by-step explanation:

Solution 1 (Algebraically):

We're given the following equation to find for all values of in the restricted domain $0\leq x \leq2\pi$ :

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

Subtract $-\sin (x)$ from both sides:

$\sin(2x)+\sin(x)=0$ .

Recall the trigonometric identity

$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$ . We can rewrite $\sin(2x)$ as $\sin(x+x)$ and use this trigonometric identity to simplify:

$\sin(2x)+\sin(x)=0,\\\\\sin(x+x)+\sin(x)=0,\\\\\sin(x)\cos(x)+\cos(x)\sin(x)+\sin(x)=0,\\\\2\cos(x)\sin(x)+\sin(x)=0,\\\\(2\cos(x)+1)(\sin(x))=0$ .

We then have two cases:

$\begin{cases}\sin(x)=0,\\2\cos(x)+1=0\end{cases}$ .

Solving, we have:

$\begin{cases}\sin(x)=0,\:x=2\pi n,\:\pi+2\pi n\:\text{for}\: n\in\mathbb{Z}\\2\cos(x)+1=0,\: x=\frac{2\pi}{3}+2\pi n,\:\frac{4\pi}{3}+2\pi n\:\text{for}\: n\in\mathbb{Z}\end{cases}$ .

However our domain is restricted to $0\leq x \leq2\pi$ . Therefore, only the following integers work for :

For $x=2\pi n,\:\pi+2\pi n,\: n\in \mathbb{Z}$ , only $n=0, \:n=1$ fit in this domain.

For $x=\frac{2\pi}{3}+2\pi n,\:\frac{4\pi}{3}+2\pi n,\:n\in\mathbb{Z}$ , only n=0 fit in this domain.

Therefore, are solutions are:

$x=2\pi(0),\:\boxed{x=0},\\x=\frac{2\pi}{3}+2\pi(0),\:\boxed{x=\frac{2\pi}{3}},\\x=\pi+2\pi(0), \:\boxed{x=\pi},\\x=2\pi(1),\:\boxed{x=2\pi}$ , where is in radians. Since the domain is given in radians ( $0\leq x \leq2\pi$ ), our answers for should be given in radians.

Solution 2 (Unit Circle):

If you have a unit circle, $\sin \theta$ is equal to the y-coordinate of the corresponding point on the unit circle. From here, you can look for angles to fit the given equation. You'll see that $0^{\circ}, \: 120^{\circ},\:180^{\circ},\: 360^{\circ}$ are the only angles that work for the given domain. Converting to radians, we get:

$x=0^{\circ}\cdot \frac{\pi}{180}=\boxed{0},\\x=120^{\circ}\cdot \frac{\pi}{180}=\boxed{\frac{2\pi}{3}},\\x=180^{\circ}\cdot \frac{\pi}{180}=\boxed{\pi},\\x=360^{\circ}\cdot \frac{\pi}{180}=\boxed{2\pi}$ .

Answer from: Quest

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

Answer from: Quest

$79500

step-by-step explanation:

75000÷100=750750×106=79500

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I’m extremely stressed and anxious again so please help me pre cal i hate how i don’t understand

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