How do you solve ${(sin (x))}^{2} = \frac{1}{25}$ $(sin(x))^2 = 1/25$ ?

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How do you solve ${(sin (x))}^{2} = \frac{1}{25}$ $(sin(x))^2 = 1/25$ ?

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Answer 1 · 2016-06-02T03:53:47

$11^{\circ} 54; 168^{\circ} 46; 191^{\circ} 54; 348^{\circ} 46$ $11^@54; 168^@46; 191^@54; 348^@46$

Explanation:

${sin}^{2} x = \frac{1}{25}$ $sin^2 x = 1/25$ --> $sin x = \pm \frac{1}{5}$ $sin x = +- 1/5$
Calculator and unit circle -->
a. $sin x = \frac{1}{5}$ $sin x = 1/5$ --> There are 2 solution arcs:
$x = 11^{\circ} 54$ $x = 11^@54$ and $x = 180 - 11^{\circ} 54 = 168^{\circ} 46$ $x = 180 - 11^@54 = 168^@46$
b. $sin x = - \frac{1}{5}$ $sin x = - 1/5$ . Two solution arcs:
$x = - 11.54$ $x = - 11.54$ or $x = 360 - 11.54 = 348.46$ $x = 360 - 11.54 = 348.46$ (co-terminal arcs)
$x = 180 + 11.54 = 191.54$ $x = 180 + 11.54 = 191.54$
Answers for $(0, 360^{\circ})$ $(0, 360^@)$ :
$11^{\circ} 54; 168^{\circ} 46; 191^{\circ} 54; 348^{\circ} 46$ $11^@54; 168^@46; 191^@54; 348^@46$

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How do you solve ${(sin (x))}^{2} = \frac{1}{25}$ $(sin(x))^2 = 1/25$ ?

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How do you solve ${(sin (x))}^{2} = \frac{1}{25}$ $(sin(x))^2 = 1/25$ ?