How do you solve # 2(sinx)^2 - 5cosx - 4 = 0# in the interval [0, 2pi]?

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Answer 1 · 2016-05-03T09:01:51

#x=(2pi)/3# or #(4pi)/3#

#2(sinx)^2-5cosx-4=0#

is equivalent to

#2(1-cos^2x)-5cosx-4=0# or

#2-2cos^2x-5cosx-4=0#

or #2cos^2x+5cosx+2=0#

or using quadratic formula

#cosx=(-5+-sqrt(5^2-4xx2xx2))/(2xx2)=(-5+-sqrt(25-16))/4#

or #cosx=(-5+-3)/4# or #cosx=-2# or #cosx=-2/4=-1/2#

But #cosx# cannot be #-2#, hence #cosx=-1/2#

or #x=2npi+-(2pi)/3#

And in interval #[0,2pi]#, #x=(2pi)/3# or #(4pi)/3#