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2 years ago

Find the equation of the problem

Mathematics

1 answer:

pogonyaev2 years ago

5 0

$6\sin^2\theta-\sin\theta=1\ \ \ -1\\\\6\sin^2\theta-\sin\theta-1=0\\\\6\sin\theta-3\sin\theta+2\sin\theta-1=0\\\\3\sin\theta(2\sin\theta-1)+1(2\sin\theta-1)=0\\\\(2\sin\theta-1)(3\sin\theta+1)=0\iff2\sin\theta-1=0\ \vee\ 3\sin\theta+1=0\\\\2\sin\theta=1\ \ |:2\ \ \vee\ \ 3\sin\theta=-1\ \ |:3\\\\\sin\theta=\dfrac{1}{2}\ \vee\ \sin\theta=-\dfrac{1}{3}\\\\$
$\theta=30^o\ \vee\ \theta=150^o\ \vee\ \theta=360^o-\sin^{-1}\left(\dfrac{1}{3}\right)\ \vee\ \theta=180^o+\sin^{-1}\left(\dfrac{1}{3}\right)$

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2 years ago

What is the square root of 5x^2 = 300

vesna_86 [32]

So, values of x are: $x=2\sqrt{15},\,\,x=-2\sqrt{15}$

Step-by-step explanation:

We need to find the square root of 5x^2 = 300

Writing in mathematical form:

$5x^2=300$

Divide both sides by 5

$\frac{5x^2}{5}=\frac{300}{5}$

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Taking square root on both sides

$\sqrt{x^2}=\sqrt{60}$

$x=\pm (\sqrt{2\times 2\times 3\times 5})$

$x=\pm (\sqrt{2^2 \times 3\times 5})$

$x=\pm (2\sqrt{3\times 5})$

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So, values of x are: $x=2\sqrt{15},\,\,x=-2\sqrt{15}$

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2 years ago

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2 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

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2 years ago