if x<-[o;360] ...

**rionariona** · February 28th 2010, 02:29 PM

if x<-[o;360] solve for x
1/sin(x-20)=-2.32

**Soroban** · February 28th 2010, 08:19 PM

Hello, rionariona!

$\text{If }x \in [0^o,\:360^o],\:\text{ solve for }x\!:\;\;\frac{1}{\sin(x-20)} \:=\:-2.32$

We have: . $\sin(x-20) \;=\;-\frac{1}{2.32}$

Then: . $x-20 \;=\;\arcsin\left(-\frac{1}{2.32}\right) \;\approx\;-25.53^o$

Hence: . $x-20 \;=\;\begin{Bmatrix}205.53^o \\ 334.47^o\end{Bmatrix}$

Therefore: . $x \;=\;\begin{Bmatrix}225.53^o \\ 354.47^o \end{Bmatrix}$

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