Questions: Square Roots Find x where (0 leq mathrmx leq 2 pi). [4 cot ^2 x=cos ^2 x+sin ^2 x-1] Enter the smaller value first.

Solution Steps

We are given the equation $4\cot^2{x} = \cos^2{x} + \sin^2{x} - 1$. Since $\cos^2{x} + \sin^2{x} = 1$, the right side simplifies to $1 - 1 = 0$. So, the equation becomes $4\cot^2{x} = 0$.

Dividing both sides by 4, we get $\cot^2{x} = 0$. Taking the square root of both sides, we have $\cot x = 0$.

Since $\cot x = \frac{\cos x}{\sin x} = 0$, we must have $\cos x = 0$ and $\sin x \ne 0$. In the interval $0 \le x \le 2\pi$, the values of x for which $\cos x = 0$ are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. For both of these values, $\sin x \ne 0$, so both are valid solutions.

Final Answer: