Solution assignment 03 Addition and subtraction formulas

Solve for $0\leq{x}\leq2\pi$ :

$2\cos^2(x)+3\sin(x)=0$

We substitute in the equation:

$\cos^2(x)=1-\sin^2(x)$

and get:

$2[1-\sin^2(x)]+3\sin(x)=0$

$2\sin^2(x)-3\sin(x)-2=0$

This is a quadratic equation in $\sin(x)$ . We use the $abc$ -formula:

$[\sin(x)]_{1,2}=\displaystyle\frac{-(-3)\pm\sqrt{(-3)^2-4.2.-2}}{2.2}=\displaystyle\frac{3\pm5}{4}$

Thus the solutions are:

$\sin(x)=2$ of $\sin(x)=-\displaystyle\frac{1}{2}$

The first equation cannot give solutions because $\sin(x)$ cannot be greater than $1$ .
For the second equation we have:

$x=\pi+\displaystyle\frac{\pi}{6}=\displaystyle\frac{7}{6}\pi$

or:

$x=2\pi-\displaystyle\frac{\pi}{6}=\displaystyle\frac{11}{6}\pi$