How do you solve #(sin(x))^2 = 1/25#?

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How do you solve #(sin(x))^2 = 1/25#?

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Answer 1 · 2016-06-02T03:53:47

#11^@54; 168^@46; 191^@54; 348^@46#

Explanation:

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

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How do you solve #(sin(x))^2 = 1/25#?

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