How do you find all solutions for #4sinthetacostheta=sqrt(3)# for the interval #[0,2pi]#?

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Answer 1 · 2014-10-24T12:34:14

4 #sin theta cos theta# = #root#3
2 #sin2theta# = #root#3
#sin 2theta#= (#root#3) /2

#2theta# = #n pi# + #(-1)^n pi /3#

#theta# = #n pi / 2# + #(-1)^n pi / 6#

For n=0, #theta#= #pi /6#

For n=1, #theta#= #pi /3#

For n=2, #theta#= #7 pi/ 6#

For n=3, #theta#= #3pi /2# - # pi/6#

For n=4, #theta# becomes greater than #2pi#.
THus the values of #theta# for n= 0, 1, 2, 3 are the solutions.