Solving trigonometric equations?

**homeylova223** · March 1st 2011, 03:22 PM

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?

**e^(i*pi)** · March 1st 2011, 03:40 PM

It's easier not to expand.

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

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

$x = \dfrac{\pi}{4} + k\pi \ \ ,\ k \in \mathbb{Z}$

**homeylova223** · March 1st 2011, 05:10 PM

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?

**topsquark** · March 1st 2011, 05:18 PM

Originally Posted by homeylova223

How would you take the arcsin in the equation I am not sure how to do it?

Arcsine (or asn or $sin^{-1}$ ) is the inverse of the sine function. So if we know that y = sin(x) then $\sin^{-1}(y) = x$ . (The angle x is written in the form $-\pi \leq x \leq \pi$ )

Thus
$sin^{-1}(sin(2x)) = 2x$

and
$sin^{-1}(1) = \frac{\pi}{2}$ because $sin \left ( \frac{\pi}{2} \right ) = 1$ .

-Dan

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