Solution assignment 07 Trigonometric equations

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

$\sin^2(x)=0.6\cos(2x)$

In the right-hand side we write for $\cos$ :

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

We could also have applied other formulas but this one fits better to the left-hand side.

$\sin^2(x)=0.6(1-2\sin^2(x))$

$2.2\sin^2(x)=0.6$

$\sin^2(x)=\displaystyle\frac{0.6}{2.2}\approx{0.27}$

$\sin(x)=\pm{0.52}$

The equation:

$\sin(x)=0.52$

has the following solution (within the domain):

$x=0.55$

or:

$x=\pi-0.55=2.59$

The equation:

$\sin(x)=-0.52$

has the solution:

$x=-0.55$

(this one does not lie in the domain and is not valid)

or:

$x=\pi-(-0.55)=3.69$

Again, this solution does not lie in the domain and is not valid.

Thus the solutions are:

$x=0.55$

or:

$x=2.59$