Question 767520
<font face="Times New Roman" size="+2">

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(a)\ +\ 1.5\sin(a)\ =\ 1]


Let *[tex \LARGE u\ =\ \sin(a)], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ +\ 1.5u\ -\ 1\ =\ 0]


Solve the quadratic, which factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(u\ +\ 2\right)\left(u\ -\ \frac{1}{2}\right)\ =\ 0]


Then substitute back, so 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(a)\ =\ -2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(a)\ =\ \frac{1}{2}]


Discard the *[tex \LARGE -2] root because the range of the *[tex \LARGE y\ =\ \sin\varphi] function is *[tex \LARGE \left\{y\ \in\ \mathbb{R}\ |\ -1\ \leq\ y\ \leq\ 1\right\}]


Using the unit circle:


<img src="http://upload.wikimedia.org/wikipedia/commons/4/4c/Unit_circle_angles_color.svg">


The value of *[tex \LARGE \sin] for an angle is the *[tex \LARGE y]-coordinate of the point of intersection of the terminal ray of the angle and the unit circle.  So find the two points in the desired interval where the *[tex \LARGE y]-coordinate equals *[tex \LARGE \frac{1}{2}]


Solve your other problem the same way, but you might want to reconsider the interval for your solutions if you are going to have any solutions at all.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>