Question

Solve sin^-1 X - cos^-1 X = sin^-1 (3x-2)?

Answer 1

@satellite73 @lgbasallote @hartnn @Shane_B @Mimi_x3

Answer 2

please write it properly
\[ sin^{-1}x - cos^{-1}x = sin^{-1}{3x-2}\]

Answer 3

Is that right?

Answer 4

yup

Answer 5

@Mimi_x3 i dont knw LATEX......

Answer 6

this one is much easier

Answer 7

\(\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}\) so
\[\cos^{-1}(x)=\frac{\pi}{2}-\sin^{-1}(x)\] and you have
\[2\sin^{-1}(x)-\frac{\pi}{2}=\sin^{-1}(3x-2)\] hmm now i am stuck

Answer 8

for some reason, this is the same as
\[\cos^{-1}(2-3x)=2\sin^{-1}(x)\] but i am not sure why
this i can solve

Answer 9

ok i got that
it is because
\[\sin^{-1}(3x-2)+\cos^{-1}(3x-2)=\frac{\pi}{2}\] and so
\[\sin^{-1}(3x-2)=\frac{\pi}{2}-\cos(3x-2)=\frac{\pi}{2}+\cos^{-1}(2-3x)\]

Answer 10

so on the right we get
\[\frac{\pi}{2}+\cos^{-1}(2-3x)\] on the left we can write
\[\sin^{-1}(x)-\cos^{-1}(x)=2\sin(x)+\frac{\pi}{2}\] since \[\cos^{-1}(x)=\frac{\pi}{2}-\sin^{-1}(x)\]
damn i messed up somewhere

Answer 11

well, if you can figure out why this says
\[2\sin^{-1}(x)=\cos^{-1}(2-3x)\] then it is easy to solve
take the cosine of both sides and you will get
\[1-2x^2=2-3x\] solve and get \(x=1\) or \(x=\frac{1}{2}\)

Answer 12

Yupp....@satellite73 u r correct