Find the value of $x$ , between $0^{0}$ and $360^{0}$ , for which: $\sin x=0.25$

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godwin_ibrahimTyro

Asked: November 25, 20202020-11-25T19:20:19+00:00 2020-11-25T19:20:19+00:00In: Calculus and Trigonometry

Find the value of $x$ , between $0^{0}$ and $360^{0}$ , for which: $\sin x=0.25$

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Find the value of $x$ , between $0^{0}$ and $360^{0}$ , for which: $\sin x=0.25$

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Sengemegz · Accepted Answer · 2020-12-07T12:14:32+00:00

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Sengemegz Moderator

2020-12-07T12:14:32+00:00Added an answer on December 7, 2020 at 12:14 pm

$\sin x=0.25$
divide through by $\sin$
$\therefore x=\arcsin (0.25)=14.5^{0}$
$\sin \theta$ is positive in the 1st and 2nd quadrants
$\Rightarrow (180-14.5)^{0}=165.5^{0}$
$\therefore x=14.5^{0} and 165.5^{0}$

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