How to solve the equation #cos^4(x)-sin^4(x)=1+sin(2x)#?

2 Answers
George C.
Jan 7, 2018

#x = npi" "# or #" "x = npi-pi/4" "# for any integer #n#

Explanation:

Given:

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

First note that the left hand side will simplify somewhat:

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

#color(white)(cos^4(x) - sin^4(x)) = cos^2(x)-sin^2(x)#

#color(white)(cos^4(x) - sin^4(x)) = cos(2x)#

So our given equation becomes:

#cos(2x) = 1+sin(2x)#

Subtracting #sin(2x)# from both sides and transposing, we find:

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

Divide both sides by #sqrt(2)# (for reasons which should become apparent) to find:

#sqrt(2)/2 = sqrt(2)/2 cos(2x) - sqrt(2)/2 sin(2x)#

#color(white)(sqrt(2)/2) = cos(pi/4) cos(2x) - sin(pi/4) sin(2x)#

#color(white)(sqrt(2)/2) = cos(2x+pi/4)#

Hence:

#2x+pi/4 = +-pi/4+2npi" "# for any integer #n#

So:

#2x = 2npi" "# or #" "2x = 2npi-pi/2#

Divide both sides by #2# to find:

#x = npi" "# or #" "x = npi-pi/4#

Cesareo R.
Jan 7, 2018

See below.

Explanation:

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

So the equivalent equation is

#cos(2x)=1+sin(2x)#

now making #cos(2x) = q#

#q=1+sqrt(1-q^2)#

now squaring both sides

#(q-1)^2=1-q^2 rArr 2q^2-2q=0# --- [1] or

#cos(2x)(cos(2x)-1)=0# with solutions

#x = {(pm pi/4+k pi)uu(k pi)}# for #k in ZZ#

but #pi/4+k pi# is not solution for the initial equation

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

This is due to the squaring operation introduced in the step #[1]# which eventually could introduce extraneous solutions.

Concluding, the solution set is

#x = {(-pi/4+k pi)uu(k pi)}# for #k in ZZ#

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