2 years ago

Please help with 1 , and 2 I really don’t understand this!

Mathematics

2 answers:

makvit [3.9K]2 years ago

8 0

Answer:

1.B

2.A

<em>Hope it helped all of you....</em>

elena-s [515]2 years ago

6 0

1. B
2. A

hope this helped

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alukav5142 [94]

Answer:

below

Step-by-step explanation:

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Round 29 to 30

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Sin^2(x/2)=sin^2x <br> Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
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$2\cos^2x-\cos x-1=0$
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The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .