How do you solve the following equation $8sinx+8=4$ in the interval [0, 2pi]?

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How do you solve the following equation $8sinx+8=4$ in the interval [0, 2pi]?

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Answer 1 · 2016-08-01T03:54:58

(7pi)/6 and (11pi)/6

Explanation:

8sin x + 8 = 4
8sin x = -4
$sin x = -4/8 = - 1/2$
Trig table gives:
$sin x = - 1/2$ --> arc $x = -pi/6$ , or $x = (11pi)/6$ (co-terminal).
The trig unit circle gives another arc x that has the same sine value:
$x = (7pi)/6.$
Answers for $(0, 2pi)$ :
$x = (7pi)/6 and x = (11pi)/6$
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How do you solve the following equation $8sinx+8=4$ in the interval [0, 2pi]?

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