Answer:
[tex]\displaystyle x=\left\{\frac{\pi}{6}+2n\pi, \frac{5\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi\right\}, n\in\mathbb{Z}[/tex]
Step-by-step explanation:
We are given the equation:
[tex]2\cos^2(x)=\sin(x)+1[/tex]
And we want to find all solutions for x.
First, we should put the equation into terms of only one trigonometric ratio.
Since we are given cos²(x), we can turn this into sine. Recall the Pythagorean Identity which states:
[tex]\sin^2(x)+\cos^2(x)=1[/tex]
Therefore:
[tex]\cos^2(x)=1-\sin^2(x)[/tex]
By substitution:
[tex]2(1-\sin^2(x))=\sin(x)+1[/tex]
Distribute:
[tex]2-2\sin^2(x)=\sin(x)+1[/tex]
Isolate the equation:
[tex]2\sin^2(x)+\sin(x)-1=0[/tex]
We can factor:
[tex](2\sin(x)-1)(\sin(x)+1)=0[/tex]
Zero Product Property:
[tex]2\sin(x)-1=0\text{ or } \sin(x)+1=0[/tex]
Solve for each case:
[tex]\displaystyle \sin(x)=\frac{1}{2}\text{ or } \sin(x)=-1[/tex]
We can use the unit circle.
sin(x) = 1/2 for every π/6 and 5π/6. So, it will continue every 2π.
sin(x) = -1 every 3π/2. And this will also continue every 2π.
Hence, our solutions are:
[tex]\displaystyle x=\left\{\frac{\pi}{6}+2n\pi, \frac{5\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi\right\}, n\in\mathbb{Z}[/tex]
Note:
If you only need the solutions within the interval [0, 2π), then it is:
[tex]\displaystyle x=\left\{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\}[/tex]
