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#

Answer link

Impact of this question

1301 views around the world

You can reuse this answer
Creative Commons License

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question