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3 years ago

PleASEE HELP!!! I HAVE TO GET AN A PLEASE

Mathematics

1 answer:

Ainat [17]3 years ago

3 0

Answer:

See below

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Yay another one of these!

1) Corr. angles postulate

2) Division prop of equality

3) Division prop of equality

Tell me if you can't use the definitions more than once

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only a few more before I advance, thanks!

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3 years ago

Cos^2 x+4sin^2 x/2=1

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$Let\ \dfrac{x}{2}=a,\ therefore\ x=2a.\\\\\cos^2x+4\sin^2\dfrac{x}{2}=\cos^22a+4\sin^2a\\\\\text{use}\ \cos2x=\sin^2x-\cos^2x\\\\=(\sin^2a-\cos^2a)^2+4\sin^2a\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(\sin^2a)^2-2(\sin^2a)(\cos^2a)+(\cos^2a)^2+4\sin^2a\\\\=\sin^4a-2\sin^2a\cos^2a+\cos^4a+4\sin^2a\\\\=\underbrace{\sin^4a+2\sin^2a\cos^2a+\cos^4a}_{(*)}-4\sin^2a\cos^2a+4\sin^2a\\\\\text{use}\ (*)\qquad(a+b)^2=a^2+2ab+b^2$

$=\underbrace{(\sin^2a)^2+2\sin^2a\cos^2a+(\cos^2a)^2}_{(*)}-4\sin^2a(\cos^2a-1)\\\\=(\sin^2a+\cos^2a)^2-4\sin^2a(\cos^2a-1)\\\\\text{use}\ \sin^2a+\cos^2a=1\to\sin^2a=\cos^2a-1\\\\=1^2-4\sin^2a(\sin^2a)=1-4\sin^4a=1-(2\sin^2a)^2$

$\cos^22a+4\sin^2a=1\\\\1-(2\sin^2a)^2=1\qquad\text{subtract 1 from both sides}\\\\-(2\sin^2a)^2=0\to2\sin^2a=0\qquad\text{divide both sides by 2}\\\\\sin^2a=0\to\sin a=0\\\\a=k\pi\ for\ k\in\mathbb{Z}\\\\\dfrac{x}{2}=k\pi\qquad\text{multiply both sides by 2}\\\\\boxed{x=2k\pi\ for\ k\in\mathbb{Z}}$

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