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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