How do you solve for the exact solutions in the interval [0,2pi] of #sin^2x-cos^2x=0#?

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Answer 1 · 2016-02-14T02:51:19

#x=45^@, 135^@, 225^@, 315^@#or
#x=pi/4, (3pi)/4, (5pi)/4, (7pi)/4#

the given #sin^2 x-cos^2 x=0#

solution by factoring using difference of two squares

#(sin x-cos x) (sin x+cos x)=0#

equate each factor to zero

#sin x-cos x=0#
#sin x=cos x#
divide both sides by #cos x#
#sin x/cos x=cos x/cos x#

#tan x=1#
#x=tan^-1 (1)#
#x=45^@, 225^@#

use the second factor

#sin x+cos x=0#
#sin x=-cos x#

divide both sides by #cos x#

#sin x/cos x=-cos x/cos x#

#tan x=-1#

#x = tan^-1(-1)#

#x=135^@, 315^@#

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