What angle is $\sin^{-1}(3/2)$? So i have this trigonometric equations: $$2\cos^2(x)+4\sin(x)+\cos(2x)=0$$
I have rewritten the expression and came up with $$(2\sin(x)-3)(1+2\sin(x))=0$$
Then i split the equation in two and got
$$\sin(x)=3/2,\quad\text{and}\quad\sin(x)=-1/2$$
Since $\sin(x)=-1/2$ is a standard angle and with respect to the period I got $x=\pi - \sin^{-1}(11\pi/6)$
The problem is that I can not figure out what angle $\sin(x)=(3/2)$.
I have tried using the Pythagorean theorem but since $\sin$ is $\text{opp}/\text{hyp}$, it tells me that something is wrong since the hypotenuse can not be shorter than the sides.
Does anyone have an idea of what I should do?

Solve
$$\sin(x)=\frac{3}{2}$$
for $x$.
Rewrite the sine function in terms of exponential function:
$$\frac{3}{2}=\sin(x)=\frac{e^{ix}-e^{-ix}}{2i};$$
multiply both sides by $2i$:
$$3i=e^{ix}-e^{-ix};$$
subtract $3i$ from both sides:
$$e^{ix}-3i-e^{-ix}=0;$$
multiply both sides by $e^{ix}$:
$$\left(e^{ix}\right)^2-3ie^{ix}-e^{-ix}e^{ix}=0;$$
rewrite $e^{-ix}e^{ix}=1$:
$$\left(e^{ix}\right)^2-3ie^{ix}-1=0;$$
substitute $u=e^{ix}$:
$$u^2-3iu-1=0;$$
multiply both sides by $4$:
$$4u^2-12iu-4=0;$$
rewrite $4u^2=\left(2u\right)^2$:
$$\left(2u\right)^2-12iu-4=0;$$
add $4+9i^2$ to both sides:
$$\left(2u\right)^2-12iu+9i^2=4+9i^2;$$
rewrite $9i^2=\left(3i\right)^2$:
$$\left(2u\right)^2-12iu+\left(3i\right)^2=4+9i^2;$$
rewrite $\left(2u\right)^2-12iu+\left(3i\right)^2=\left(2u-3i\right)^2$:
$$\left(2u-3i\right)^2=4+9i^2;$$
rewrite $4+9i^2=4-9=-5$:
$$\left(2u-3i\right)^2=-5;$$
take the square roots of both sides:
$$2u-3i=\pm i\sqrt{5};$$
add $3i$ to both sides:
$$2u=3i\pm i\sqrt{5};$$
factor $i$ out:
$$2u=i\left(3\pm\sqrt{5}\right);$$
divide both sides by $2$:
$$u=\frac{i}{2}\left(3\pm\sqrt{5}\right);$$
substitute back $u=e^{ix}$:
$$e^{ix}=\frac{i}{2}\left(3\pm\sqrt{5}\right);$$
rewrite $\frac{i}{2}\left(3\pm\sqrt{5}\right)=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)}$:
$$e^{ix}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)};$$
since $1=e^{2i\pi c_1}$ for arbitrary integer $c_1$, multiply the RHS by $e^{2i\pi c_1}$:
$$e^{ix}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)}e^{2i\pi c_1};$$
rewrite $e^{\log\left(3i/2\pm i\sqrt{5}/2\right)}e^{2i\pi c_1}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)+2i\pi c_1}$:
$$e^{ix}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)+2i\pi c_1};$$
eliminate exponentials:
$$ix=\log\left(\frac{3\pm\sqrt{5}}{2}i\right)+2i\pi c_1;$$
divide both sides by $i$:
$$x=\frac{1}{i}\log\left(\frac{3\pm\sqrt{5}}{2}i\right)+2\pi c_1;$$
rewrite $1/i=-i$:
$$x=-i\log\left(\frac{3\pm\sqrt{5}}{2}i\right)+2\pi c_1;$$
rewrite $\log\left(\frac{3\pm\sqrt{5}}{2}i\right)=\log\left(\frac{3\pm\sqrt{5}}{2}\right)+\frac{i\pi}{2}$:
$$x=-i\left(\log\left(\frac{3\pm\sqrt{5}}{2}\right)+\frac{i\pi}{2}\right)+2\pi c_1;$$
expand:
$$x=-i\log\left(\frac{3\pm\sqrt{5}}{2}\right)-\frac{i^2\pi}{2}+2\pi c_1;$$
rewrite $i^2=-1$:
$$x=\frac{\pi}{2}-i\log\left(\frac{3\pm\sqrt{5}}{2}\right)+2\pi c_1.$$
From the whole set of solutions, the principal inverse sine value is delivered by setting $c_1=0$ and choosing the positive argument of $\log$:
$$\arcsin\left(\frac{3}{2}\right)=\boxed{\frac{\pi}{2}-i\log\left(\frac{3+\sqrt{5}}{2}\right)},$$
where $\log$ is the natural logarithm and $i$ is the imaginary unit.
