Question

Find the all solutions of this equation in the interval [0,2pi)2cos(x)+sqrt2=0

Answer 1

The given equation is:

$2\cos x+\sqrt[]{2}=0$

Move the constant to the right, changing its sign:

$\Rightarrow2\cos x=-\sqrt[]{2}$

Divide both sides of the equation by 2:

$\begin{gathered} \frac{2\cos x}{2}=\frac{-\sqrt[]{2}}{2} \\ \Rightarrow\cos x=\frac{-\sqrt[]{2}}{2} \end{gathered}$

Since cos t = cos(2π-t), it follows that the equation has two solutions:

$\cos x=-\frac{\sqrt[]{2}}{2};\; \cos \text{ (2}\pi-x)=-\frac{\sqrt[]{2}}{2}$

Consider the first equation:

$\begin{gathered} \cos x=-\frac{\sqrt[]{2}}{2} \\ Use\text{ the inverse trigonometry function to isolate x:} \\ \Rightarrow x=\arccos (-\frac{\sqrt[]{2}}{2}) \\ \Rightarrow x=\frac{3\pi}{4} \\ \end{gathered}$

Since the cosine function is periodic, add the period to the solution:

$\Rightarrow x=\frac{3\pi}{4}+2\pi k;\; k\in Z$

Consider the second equation and apply the same procedure as before:

$\begin{gathered} \; \cos \text{ (2}\pi-x)=-\frac{\sqrt[]{2}}{2} \\ \Rightarrow2\pi-x=\frac{3\pi}{4}+2\pi k \\ \Rightarrow-x=\frac{3\pi}{4}-2\pi+2\pi k \\ \Rightarrow-x=-\frac{5\pi}{4}+2\pi k \\ \Rightarrow x=\frac{5\pi}{4}-2\pi k \end{gathered}$

Since k is an integer, then -2πk=2πk.

$\Rightarrow x=\frac{5\pi}{4}+2\pi k$

Hence the solutions are:

$\begin{cases}x=\frac{3\pi}{4}+2\pi k \\ x=\frac{5\pi}{4}+2\pi k\end{cases}k\in Z$

Notice that the solutions are required to be in [0,2π).

Hence, substitute integer values of k into the solutions and find values of x that fall in the given interval of solutions.

$\begin{gathered} x=\frac{3\pi}{4}+2\pi k;k=0 \\ \Rightarrow x=\frac{3\pi}{4}+2\pi(0)=\frac{3\pi}{4} \end{gathered}$

Other values of k will not fall in the required interval for this solution.

Check for the second solution:

$\begin{gathered} x=\frac{5\pi}{4}+2\pi k;k=0 \\ \Rightarrow x=\frac{5\pi}{4}+2\pi(0)=\frac{5\pi}{4} \end{gathered}$

Other values for other integer values do not fall in the interval.

Hence, the solutions are:

$x=\frac{3\pi}{4},\frac{5\pi}{4}$