How do you solve #(2sintheta-1)^4=0#?

1 Answer

Aug 31, 2016

#θ = pi/6 + 2kpi# or #θ = 5/6pi + 2kpi#, #k in Z#

Explanation:

if #(2sinθ−1)^4=0#
then #2sinθ−1 = 0#
thus #sinθ = 1/2#
elementary solution #θ_0# =#pi/6#
adding period
#θ = pi/6 + 2kpi# or #θ = 5/6pi + 2kpi#, #k in Z#

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How do you solve #(2sintheta-1)^4=0#?

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