How do you solve #4sin^2x+5=6#?

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Answer 1 · 2016-09-27T17:09:05

#x=npi+-pi/6#, where #n# is an integer

#4sin^2x+5=6#

#hArr4sin^x+5-6=0# or

#4sin^x-1=0#

or #(2sinx)^2-1^2=0#

or #(2sinx+1)(2sinx-1)=0#

so either #2sinx=-1# or #2sinx=1#

i.e. #sinx=-1/2# or #sinx=1/2#

Now #sinx=1/2#, when #x=pi/6# or #(5pi)/6#

and #sinx=-1/2#, when #x=-pi/6# or #-(5pi)/6#

as #(5pi)/6=pi-pi/6#, we can summarize result, in general, as

#x=npi+-pi/6#, where #n# is an integer