How do you solve $sin^2 x - cos^2 x=0$ for x in the interval [0,2pi)?

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Answer 1 · 2016-03-10T23:20:27

$x in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}$

Applying the identity $cos^2(x)-sin^2(x) = cos(2x)$ we have

$sin^2(x)-cos^2(x) = -cos(2x)$

In general,

$cos(u) = 0 <=> u = (npi)/2$ for some $n in ZZ$

Thus we have

$sin^2(x) - cos^2(x) = 0$

$=>-cos(2x) = 0$

$=> 2x = (npi)/2$ for $n in ZZ$

$=> x = (npi)/4$ for $n in ZZ$

Restricting our values to the interval $[0, 2pi]$ gives our final result:

$x in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}$

How do you solve sin^2 x - cos^2 x=0sin^2 x - cos^2 x=0 for x in the interval [0,2pi)?