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Anuta_ua [19.1K]

3 years ago

7

give the most general solutions to equations. . 2sinxcosx-sin(2x)cos(2x)=0. . A. simplify the first expression using double angl

e identity for sine.. B. factor the left side of equatioin. C. solve the factored equation

1 answer:

Zinaida [17]3 years ago

5 0

2sinxcosx - sin(2x)cos(2x) = 0

<span>Part I </span>
<span>The double angle identity for sine states that sin(2x) = 2sinxcosx </span>
<span>Thus we get: </span>
<span>sin(2x) - sin(2x)cos(2x) = 0 </span>

<span>Part II </span>
<span>sin(2x)(1 - cos(2x)) = 0 </span>

<span>Part III </span>
<span>Either sin(2x) = 0 or </span>
<span>1 - cos(2x) = 0 </span>
<span>=> cos(2x) = 1 </span>

<span>For sin(2x) = 0, this is true for </span>
<span>2x = n(pi) where n = 0, 1, 2, .... </span>
<span>x = n(pi/2) </span>

<span>For cos(2x) = 1, this is true for </span>
<span>2x = n(pi) where n = 0, 2, 4, .... </span>
<span>x = n(pi/2)
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

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2 years ago

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The correct answer is D.

This is because a converse statement switches the places of the hypothesis and conclusion.

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3 years ago

Alex’s paycheck for the last two weeks totaled $239.64. He owes the cell phone company $45.87. How much money does he have left?

enyata [817]

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3 years ago

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Jlenok [28]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

In problems 16-20, calculate the expected price in the year 2008 if you assume that there was a consistent 5% inflation rate and

Lena [83]

16. 72236.03
17. <span>2.87
</span>18. 2.36
19. 9.29
20. 1.65
The exponential growth formula is as follows
$y=a(1+r)^{x}$ :
a is initial value
(<span>$27,225 17. gallon of gas, $1.08 18. dozen eggs, $0.89 19. movie admission, $3.50 20. mcdonalds hamburger, $0.62)
r is rate of increase (0.05)
x is time (2008-1988 = 20 years)
to solve plug in values for each one.
</span> $y=27225(1+0.05)^{20} = 72236.03$

$1.08(1+0.05)^{20} = 2.87$

$.89(1+0.05)^{20} = 2.36$

$3.50(1+0.05)^{20} = 9.29$

$.62(1+0.05)^{20} = 1.65$

<span>

</span>

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3 years ago