Question

cos^2x=Sin^2x

find x

Answer 1

I know no solution, just made this up, can you help me with literature?

Answer 2

http://openstudy.com/study#/updates/529613bbe4b0ec64e3aeab91

Answer 3

http://openstudy.com/study#/updates/529613bbe4b0ec64e3aeab91

Answer 4

1-sin^2x=sin^2x
1=2sin^2x
1/2=sin62x
sinx=1/4

Answer 5

sin^(-1) 1/4

Answer 6

that's a rude way to attract people to literature qestion, sorry.

Answer 7

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

Answer 8

you don't need all this.
\[1-Sin^2x=Sin^2x~~~~~~~~~~~~~~~->~~~~~~~~~~~~1=2Sin^2x\]\[Sin ^{-1}(1/4)\]