Solution assignment 10 Trigonometric equations

Solve for $0\leq{x}\leq{2\pi}$ :

$6\sin^2(x)-5\sin(x)+1=0$

We substitute:

$y=\sin(x)$

in the equation and we get:

$6y^2-5y+1=0$

This is a quadratic equation and the solutions are:

$y_{1,2}=\displaystyle\frac{-(-5)\pm\sqrt{(-5)^2-4.6.1}}{12}=\displaystyle\frac{5\pm1}{12}$

$y_1=\displaystyle\frac{1}{2}$ or $y_2=\displaystyle\frac{1}{3}$

Now we have to solve the equations:

$\sin(x_1)=\displaystyle\frac{1}{2}$

$\sin(x_2)=\displaystyle\frac{1}{3}$

The first equation yields:

$x_1=\displaystyle\frac{\pi}{6}=0.52$

or:

$x_1=\displaystyle\frac{5\pi}{6}=2.61$

The second solution can be calculated using a calculator. We find approximately:

$x_2=0.34$

or:

$x_2=2.80$

Thus the equation has four solutions in the given interval:

$0.34; 0.52; 2.61; 2.80$