Question

solve the equation on the interval [0, 2pi] sin^2(x)-cos^2(x)=0

Answer 1

\(\bf sin^2(x)-cos^2(x)=0 \implies -[cos^2(x)-sin^2(x)]=0\\
\color{blue}{cos(2x) = cos^2(x)-sin^2(x)\\
cos(2x) = 2cos^2(x)-1}\\
\implies \color{green}{cos^2(x)-sin^2(x) = 2cos^2(x)-1}\\
-[2cos^2(x)-1] = 0 \implies 1-2cos^2(x) = 0\)

and I gather you can take it from there

Answer 2

move cos^2 over to right side
\[\sin^{2} x = \cos^{2} x\]
\[\frac{\sin^{2} x}{\cos^{2} x} = 1\]
\[\tan^{2} x = 1\]
\[\tan x = \pm 1\]

Answer 3

as you can see there is more than 1 way to get answers

Answer 4

\[\sin ^{2}x-\cos ^{2}x=0,2\sin ^{2}x=2\cos ^{2}x,1-\cos 2x=1+\cos 2x\]
\[\cos 2x=0=\cos \left( \frac{ 2n+1 }{2 } \right)\pi ,n=0,1,2,......\]
\[2x=\frac{ \left( 2n+1 \right) }{2 }\pi,n=0,1,2,....\]
\[x=\frac{ \left( 2n+1 \right) }{ 4 }\pi ,n=0,1,2,3,......\]

Answer 5

so should the answers be pi/4, 3pi/4, 5pi/4, and 7pi/4

Answer 6

@dumbcow

Answer 7

correct...all the multiples of pi/4 are solutions

Answer 8

ok thank you!!