<img src="https://tex.z-dn.net/?f=%20express%20%5C%3A%20%20%5C%3A%20%5Csin%282x%29%20%20-%20%20%5Csqrt%7B3%7D%20%20%5Ccos%282x%2

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3 years ago

9%20as%20%5C%3A%20%5C%3A%20...%5Cgamma%20%20%5Csin%282x%20-%20%20%5Calpha%20%29.%20%20%5Calpha%20%5C%3A%20%20is%20%5C%3A%20%20%5C%5C%20%20given%20%5C%3A%20%5C%3A%20%20as%20%5C%3A%20%20%5C%3A%20%20%280%20%3C%20%20%5Calpha%20%20%3C%20%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%29%20%5C%3A%20and%20%5C%3A%20%20%5C%3A%20%20%5Cgamma%20%28%20%3E%200%29.%20%5C%3A%20find%20%5C%3A%20%20%5C%3A%20%20%5Calpha%20%5C%3A%20%20%5C%3A%20%20and%20%20%5C%3A%20%20%20%5C%3A%20%5Cgamma%20...%20%5C%5C%20%20%5C%5C%20find%20%5C%3A%20the%20%5C%3A%20solutions%20%5C%3A%20for%20%5C%3A%20%20%5Csin%282x%29%20%20-%20%20%5Csqrt%7B3%7D%20%20%5Ccos%282x%29%20%20%3D%201%20%5C%3A%20between%20...%5C%3A%20%20-%20%5Cpi%20%5Cleqslant%20x%20%5Cleqslant%20%5Cpi" id="TexFormula1" title=" express \: \: \sin(2x) - \sqrt{3} \cos(2x) as \: \: ...\gamma \sin(2x - \alpha ). \alpha \: is \: \\ given \: \: as \: \: (0 < \alpha < \frac{\pi}{2} ) \: and \: \: \gamma ( > 0). \: find \: \: \alpha \: \: and \: \: \gamma ... \\ \\ find \: the \: solutions \: for \: \sin(2x) - \sqrt{3} \cos(2x) = 1 \: between ...\: - \pi \leqslant x \leqslant \pi" alt=" express \: \: \sin(2x) - \sqrt{3} \cos(2x) as \: \: ...\gamma \sin(2x - \alpha ). \alpha \: is \: \\ given \: \: as \: \: (0 < \alpha < \frac{\pi}{2} ) \: and \: \: \gamma ( > 0). \: find \: \: \alpha \: \: and \: \: \gamma ... \\ \\ find \: the \: solutions \: for \: \sin(2x) - \sqrt{3} \cos(2x) = 1 \: between ...\: - \pi \leqslant x \leqslant \pi" align="absmiddle" class="latex-formula">
can someone help me with this plzz

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

7 0

Answer:

α = π/3

γ = 2

x = -3π/4, -5π/12, π/4, 7π/12

Step-by-step explanation:

Easiest way to do this is in reverse.

γ sin(2x − α)

Angle sum/difference formula:

γ (cos α sin(2x) − sin α cos(2x))

Distribute:

γ cos α sin(2x) − γ sin α cos(2x)

Matching the coefficients:

γ cos α = 1

γ sin α = √3

Solve the system of equations. Divide to eliminate γ:

tan α = √3

α is between 0 and π/2, so:

α = π/3

γ = 2

sin(2x) − √3 cos(2x) = 1

Using the identity from before:

2 sin(2x − π/3) = 1

Solving:

sin(2x − π/3) = 1/2

2x − π/3 = π/6 + 2kπ or 5π/6 + 2kπ

2x = π/2 + 2kπ or 7π/6 + 2kπ

x = π/4 + kπ or 7π/12 + kπ

x is between -π and π, so:

x = -3π/4, -5π/12, π/4, 7π/12

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harkovskaia [24]

Answer:

$\rm x = \dfrac{2e - 1}{ {e}^{2} }$

Step-by-step explanation:

$\rm Solve \: for \: x: \\ \rm \longrightarrow ex + {e}^{ - 1} = 2 \\ \\ \rm \longrightarrow e x + \dfrac{1}{e} = 2 \\ \\ \rm Subtract \: \dfrac{1}{e} \: from \: both \: sides: \\ \rm \longrightarrow e x + \dfrac{1}{e} - \dfrac{1}{e} = 2 - \dfrac{1}{e} \\ \\ \rm \longrightarrow ex = \dfrac{2e}{e} - \dfrac{1}{e} \\ \\ \rm \longrightarrow ex = \dfrac{2e - 1}{e} \\ \\ \rm Divide \: both \: sides \: by \: e: \\ \rm \longrightarrow \dfrac{ex}{e} = \dfrac{2e - 1}{e \times e} \\ \\ \rm \longrightarrow x = \dfrac{2e - 1}{ {e}^{2} }$

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3 years ago

Which ordered pair identifies a point in Quadrant IV? A) (2, 5) B) (0, 1) C) (-1, -1) D) (12, -5)

MariettaO [177]

Correct Answer:
Option D (12, -5)

In Quadrant IV, the x-component of the point is positive and the y-component is negative. From the given points, option D contains the point with a positive x-component i.e.12 and negative y-component i.e -5. Therefore, this point lies in Quadrant IV.

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3 years ago

H function is even, odd, or neither. b. g(x) = x² - 2 Is it odd or even

o-na [289]

$g(x) = x^2 - 2 \text{ is even function }$

Solution:

Given that,

$g(x) = x^2 - 2$

We have to find whether the above function is odd or even

If a function is: y = f(x)

If f(-x) = f(x), the function is even

If f(-x) = - f(x), the function is odd

Which is,

$\mathrm{Even\:Function:\:\:A\:function\:is\:even\:if\:}f\left(-x\right)=f\left(x\right)\mathrm{\:for\:all\:}x\in \mathbb{R}\\\\\mathrm{Odd\:Function:\:\:A\:function\:is\:odd\:if\:}f\left(-x\right)=-f\left(x\right)\mathrm{\:for\:all\:}x\in \mathbb{R}$

From given,

$g(x) = x^2 - 2$

Replace x with -x

$g(-x) = (-x)^2 - 2\\\\g(-x) = x^2 - 2$

Therefore,

$g(x) = g(-x)$

Thus the function g(x) is even

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2 years ago

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lyudmila [28]

Answer:

A. In a binomial distribution, the value ofx represents the number of successes in n trials, while in a geometric distribution, the value ofx represents the first trial that results in a success.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

The most important difference is that in the binomial distribution, the value of x represents the successes in n trials.

And by the other hand in the geometric distribution, x represents the number of failures before you get a success in a series of Bernoulli trials.

So then the best answer for this case is:

A. In a binomial distribution, the value of x represents the number of successes in n trials, while in a geometric distribution, the value ofx represents the first trial that results in a success.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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