Can anybody help me with this equation? I am staring to think there is a mistake there....
I tried to factor it, but it does not seem right...
2sin^2(x) - 6sin(x)- 3=0
Exact solutions:
Let $\displaystyle x_1 = \frac{3 - \sqrt{15}}{2}$. Then $\displaystyle \sin (x) = x_1 \Rightarrow x = \sin^{-1} (x_1) + 2 n \pi$ or $\displaystyle x = \sin^{-1} (-x_1) + \pi + 2 n \pi$ (using the symmetry of the unit circle) where n is an integer.
Approximate solutions:
Use a calculator to get a decimal approximation of $\displaystyle \sin^{-1} \left( \frac{3 - \sqrt{15}}{2} \right)$ and then proceed in the usual way.
So, I calculated it and got appx: -26
I must be missing something, because I really do not know how to proceed from here...
I know how to use unit circle, but what is value of x, when sinx=-26??
I am veeery confused now....