How do you solve #2cos^2(x)+ 4sin^2(x)=3#?

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Answer 1 · 2015-05-11T19:41:35

Using #sin^2(x)+cos^2(x) = 1#, we can express the equation as:

#3 = 2cos^2(x)+4sin^2(x) = 2(1-sin^2(x))+4sin^2(x)#

#=2-2sin^2(x)+4sin^2(x)#

#=2+2sin^2(x)#

Subtract 2 from both ends to get:

#2sin^2(x)=1#

Divide both sides by 2 to get:

#sin^2(x)=1/2#

Again, since #sin^2(x)+cos^2(x) = 1#, we also have:

#cos^2(x)=1-sin^2(x)=1-1/2=1/2#

Hence #sin(x)=+-sqrt(2)/2# and #cos(x)=+-sqrt(2)/2#

This is true for #x = pi/4+(n pi)/2# for all integer values of #n#.