2 years ago

No links or i will report

Mathematics

1 answer:

yanalaym [24]2 years ago

6 0

Answer:

The answers go in the following order:

33.6

68.4

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Solve for x: 2sin^2(x) -sin(x)=1 over the interval 0 to 2pi

Zepler [3.9K]

$2\sin^2x-\sin x=1\ \ \ \ |subtract\ 1\ from\ both\ sides\\\\2\sin^2x-\sin x-1=0\ \ \ \ |subtitute\ \sin x=t\ where\ t\in[-1;\ 1]\\\\2t^2-t-1=0\\\\2t^2-2t+t-1=0\\\\2t(t-1)+1(t-1)=0\\\\(t-1)(2t+1)=0\iff t-1=0\ or\ 2t+1=0\\\\t=1\ or\ 2t=-1\to t=-\dfrac{1}{2}\\\\therefore\\\sin x=1\ or\ \sin x=-\dfrac{1}{2}\\\\Answer:\boxed{t\in\left\{\dfrac{\pi}{2};\ \dfrac{7\pi}{6};\ \dfrac{11\pi}{6}\right\}}$

6 0

3 years ago

NEED HELP ASAP WILL GIVE BRAINLY IF CORRECT

yanalaym [24]

Answer:

$2x - 9y = 1 \ \: \\ \: is \: \: the \: \: correct \: \: answer \: \: of \: \: this \: \: question$