find all solutions of the equation in the interval 0, 2π). 2sin²x - 1 = 0 write your answer in radians in…

Explanation:

Step1: Isolate $\sin^{2}x$

Add 1 to both sides of the equation $2\sin^{2}x - 1=0$ and then divide by 2. $2\sin^{2}x=1$, so $\sin^{2}x=\frac{1}{2}$.

Step2: Solve for $\sin x$

Take the square - root of both sides. $\sin x=\pm\frac{\sqrt{2}}{2}$.

Step3: Find solutions for $\sin x=\frac{\sqrt{2}}{2}$ in $[0,2\pi)$

We know that $\sin x=\frac{\sqrt{2}}{2}$ when $x = \frac{\pi}{4}$ and $x=\frac{3\pi}{4}$ in the interval $[0,2\pi)$ since the sine function is positive in the first and second quadrants.

Step4: Find solutions for $\sin x=-\frac{\sqrt{2}}{2}$ in $[0,2\pi)$

We know that $\sin x =-\frac{\sqrt{2}}{2}$ when $x=\frac{5\pi}{4}$ and $x=\frac{7\pi}{4}$ in the interval $[0,2\pi)$ since the sine function is negative in the third and fourth quadrants.

Answer:

$\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$