3 years ago

I need help on the math question look at pic

Mathematics

1 answer:

frutty [35]3 years ago

5 0

Answer:

9 is answer . . . . . . . ........

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What is an obtuse angle?

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2 years ago

Solve the equation on the<br> interval [0, 27r).<br> 4(sin x)2 - 2 = 0

Ket [755]

$4[sin(x)]^2 - 2 = 0\implies 4[sin(x)]^2=2\implies [sin(x)]^2=\cfrac{2}{4}\implies [sin(x)]^2=\cfrac{1}{2} \\\\\\ sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies sin^{-1}[sin(x)]=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)\implies x=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)$

$x=sin^{-1}\left( \pm\cfrac{\sqrt{1}}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{1}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill x=\cfrac{\pi }{4}~~,~~\cfrac{3\pi }{4}~~,~~\cfrac{5\pi }{4}~~,~~\cfrac{7\pi }{4}~\hfill$