What are all the solutions between 0 and 2π for #2cos^2 = sinx+1#?

1 Answer
Mar 29, 2016

Answer:

#x in {pi/6, (5pi)/6, (3pi)/2}#

Explanation:

Assuming the equation was meant to be:
#color(white)("XXX")2cos^2(color(red)(x))=sin(x)+1#

Using the relationship:
#color(white)("XXX")cos^2(x)+sin^2(x)=1#
#2 cos^2(x)=sin(x)+1#

#rarr 2(1-sin^2(x))=sin(x)+1#

#rarr 2sin^2(x)+sin(x)-1=0#

#rarr (2sin(x)-1)(sin(x)+1)=0#

#rarr sin(x)=1/2color(white)("XXX")orcolor(white)("XXX")sin(x)=-1#
#rarr x=pi/6 or (5pi)/6color(white)("XXXXXXX")rarr x=(3pi)/2#
#color(white)("XXXXXXXXXX")#(for #x in[0,2pi]#)