Question

How do I find all solutions for$2{\cos }^{2}x-1=0$ on $[0,2\pi ]$?

Answer 1

Find all solutions for$2{\cos }^{2}x-1=0$ on $[0,2\pi ]$.

Add $1$ on both sides of the equation:

$$\begin{gathered} 2{\cos }^{2}x-1+1=0+1 \\ \Rightarrow 2{\cos }^{2}x=1 \end{gathered}$$

Divide $2$ on both sides:

$$\begin{gathered} \frac{2{\cos }^{2}x}{2}=\frac{1}{2} \\ \Rightarrow {\cos }^{2}x=\frac{1}{2} \end{gathered}$$

Take square roots on both sides:

$$\begin{gathered} \cos \left(x\right)=\pm \frac{1}{\sqrt{2}} \\ x=\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}\frac{7\mathrm{\pi }}{4} \end{gathered}$$

Hence, all solutions for$2{\cos }^{2}x-1=0$ on $[0,2\pi ]$ is $\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}\frac{7\mathrm{\pi }}{4}$.