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konstantin123 [22]
4 years ago
11

Not a hard question i just don't get it. I really need some help here, its the final

Mathematics
1 answer:
Greeley [361]4 years ago
5 0

Start with

\dfrac{1}{\sin^2(x)}=4

Invert both sides:

\sin^2(x) = \dfrac{1}{4}

Consider the square root of both sides with double sign:

\sin(x) = \pm\sqrt{\dfrac{1}{4}} = \pm\dfrac{1}{2}

We have

\sin(x) = \dfrac{1}{2} \iff x = \dfrac{\pi}{6}\ \lor\ x = \dfrac{5\pi}{6}

and

\sin(x) = -\dfrac{1}{2} \iff x = \dfrac{7\pi}{6}\ \lor\ x = \dfrac{11\pi}{6}

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4 years ago
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