Answer:
[tex]\displaystyle x=\left\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\right\}[/tex]
Step-by-step explanation:
We are given the equation:
[tex](\sin(x)-\cos(x))^2=1+\cos(x)[/tex]
And we want to find all solutions for the equation within the interval [0, 2π).
First, we can expand. This yields:
[tex]\sin^2(x)-2\sin(x)\cos(x)+\cos^2(x)=1+\cos(x)[/tex]
From the Pythagorean Identity, we know that:
[tex]\sin^2(x)+\cos^2(x)=1[/tex]
Therefore:
[tex]-2\sin(x)\cos(x)+1=1+\cos(x)[/tex]
Simplify:
[tex]-2\sin(x)\cos(x)=\cos(x)[/tex]
Subtract cos(x) from both sides*:
[tex]-2\sin(x)\cos(x)-\cos(x)=0[/tex]
Factor:
[tex]\cos(x)\left(-2\sin(x)-1)=0[/tex]
Zero Product Property:
[tex]\cos(x)=0\text{ or } -2\sin(x)-1=0[/tex]
Solve for each case:
[tex]\displaystyle \cos(x)=0\text{ or } \sin(x)=-\frac{1}{2}[/tex]
Use the unit circle. So, our solutions are:
[tex]\displaystyle x=\left\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\right\}[/tex]
*Note that we should not simply divide both sides by cos(x) to acquire -2sin(x) = 1. This is because we do not know what the value of x is, and so one or may values of x may result in cos(x) = 0, and we cannot divide by 0. Hence, we need to subtract and factor and utilize the Zero Product Property.
