How do you find the general solutions for #sin2x=sqrt3/2#?

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How do you find the general solutions for #sin2x=sqrt3/2#?

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Answer 1 · 2015-08-12T15:51:02

Solve #sin 2x = sqrt3/2#

Ans: #pi/3# and #pi/6#

Explanation:

Trig table gives -> #sin 2x = sqrt3/2# --> #2x = (2pi)/3# --> #x = pi/3#
Trig unit circle gives another arc 2x that has the same sine value:
#2x = pi - (2pi)/3 = pi/3# --> #x = pi/6.#
Answer:
#x = pi/3 + 2kpi#
#x = pi/6 + 2kpi#

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How do you find the general solutions for #sin2x=sqrt3/2#?

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