Can anyone help me solve these two problems
sin2x-1=0
2sinx cox-1=0 Then I am not sure how to proceed
sinxtanx-sinx=0
sinx (sinx/cosx)- sinx
Sinx^2/ cosx -sinxcosx/cosx I am not sure if I am doing this right?
Arcsine (or asn or $\displaystyle sin^{-1}$ ) is the inverse of the sine function. So if we know that y = sin(x) then $\displaystyle \sin^{-1}(y) = x$. (The angle x is written in the form $\displaystyle -\pi \leq x \leq \pi$ )
Thus
$\displaystyle sin^{-1}(sin(2x)) = 2x$
and
$\displaystyle sin^{-1}(1) = \frac{\pi}{2}$ because $\displaystyle sin \left ( \frac{\pi}{2} \right ) = 1$.
-Dan