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?(Thinking)

Printable View

- Mar 1st 2011, 03:22 PMhomeylova223Solving trigonometric equations?
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?(Thinking) - Mar 1st 2011, 03:40 PMe^(i*pi)
It's easier not to expand.

Add 1 to both sides: $\displaystyle \sin(2x) = 1$

Take the arcsin of both sides: $\displaystyle 2x = \dfrac{\pi}{2}$

$\displaystyle x = \dfrac{\pi}{4} + k\pi \ \ ,\ k \in \mathbb{Z}$ - Mar 1st 2011, 05:10 PMhomeylova223
Thanks for your help however the second function I am still having problem solving.

I narrowed it down to

sinx^2-sinx=0 factor sin

sinx(sinx-1)=0

sinx=0 sinx=1

x=0, 90,180,270,360 degrees? - Mar 1st 2011, 05:18 PMtopsquark
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